Annals of Mathematics, 161 (2005), 1423–1485
The Lyapunov exponents of generic
volume-preserving and symplectic maps
By Jairo Bochi and Marcelo Viana*
To Jacob Palis, on his 60th birthday, with friendship and admiration.
Abstract
We show that the integrated Lyapunov exponents of C 1 volume-preserving
diffeomorphisms are simultaneously continuous at a given diffeomorphism only
if the corresponding Oseledets splitting is trivial (all Lyapunov exponents are
equal to zero) or else dominated (uniform hyperbolicity in the projective bundle) almost everywhere.
We deduce a sharp dichotomy for generic volume-preserving diffeomorphisms on any compact manifold: almost every orbit either is projectively
hyperbolic or has all Lyapunov exponents equal to zero.
Similarly, for a residual subset of all C 1 symplectic diffeomorphisms on
any compact manifold, either the diffeomorphism is Anosov or almost every
point has zero as a Lyapunov exponent, with multiplicity at least 2.
Finally, given any set S ⊂ GL(d) satisfying an accessibility condition, for a
residual subset of all continuous S-valued cocycles over any measure-preserving
homeomorphism of a compact space, the Oseledets splitting is either dominated
or trivial. The condition on S is satisfied for most common matrix groups and
also for matrices that arise from discrete Schrödinger operators.
1. Introduction
Lyapunov exponents describe the asymptotic evolution of a linear cocycle
over a transformation: positive or negative exponents correspond to exponential growth or decay of the norm, respectively, whereas vanishing exponents
mean lack of exponential behavior.
*Partially supported by CNPq, Profix, and Faperj, Brazil. J.B. thanks the Royal Institute of Technology for its hospitality. M.V. is grateful for the hospitality of Collège de
France, Université de Paris-Orsay, and Institut de Mathématiques de Jussieu.
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JAIRO BOCHI AND MARCELO VIANA
In this work we address two basic, a priori unrelated problems. One is to
understand how frequently Lyapunov exponents vanish on typical orbits. The
other, is to analyze the dependence of Lyapunov exponents as functions of the
system. We are especially interested in dynamical cocycles, i.e. those given
by the derivatives of conservative diffeomorphisms, but we discuss the general
situation as well.
Several approaches have been proposed for proving existence of nonzero
Lyapunov exponents. Let us mention Furstenberg [14], Herman [16], Kotani
[17], among others. In contrast, we show here that vanishing Lyapunov exponents are actually very frequent: for a residual (dense Gδ ) subset of all
volume-preserving C 1 diffeomorphisms, and for almost every orbit, all
Lyapunov exponents are equal to zero or else the Oseledets splitting is dominated. This extends to generic continuous S-valued cocycles over any transformation, where S is a set of matrices that satisfy an accessibility condition,
for instance, a matrix group G that acts transitively on the projective space.
Domination, or uniform hyperbolicity in the projective bundle, means
that each Oseledets subspace is more expanded/less contracted than the next,
by a definite uniform factor. This is a very strong property. In particular,
domination implies that the angles between the Oseledets subspaces are bounded
from zero, and the Oseledets splitting extends to a continuous splitting on the
closure. For this reason, it can often be excluded a priori:
Example 1. Let f : S 1 → S 1 be a homeomorphism and µ be any invariant
ergodic measure with supp µ = S 1 . Let N be the set of all continuous A : S 1 →
SL(2, R) nonhomotopic to a constant. For a residual subset of N , the Lyapunov
exponents of the corresponding cocycle over (f, µ) are zero. That is because
the cocycle has no invariant continuous subbundle if A is nonhomotopic to a
constant.
These results generalize to arbitrary dimension the work of Bochi [4],
where it was shown that generic area-preserving C 1 diffeomorphisms on any
compact surface either are uniformly hyperbolic (Anosov) or have no hyperbolicity at all; both Lyapunov exponents equal zero almost everywhere. This
fact was announced by Mañé [19], [20] in the early eighties.
Our strategy is to tackle the higher dimensional problem and to analyze
the dependence of Lyapunov exponents on the dynamics. We obtain the following characterization of the continuity points of Lyapunov exponents in the
space of volume-preserving C 1 diffeomorphisms on any compact manifold: they
must have all exponents equal to zero or else the Oseledets splitting must be
dominated, over almost every orbit. This is similar for continuous linear cocycles over any transformation, and in this setting the necessary condition is
known to be sufficient.
1425
LYAPUNOV EXPONENTS
The issue of continuous or differentiable dependence of Lyapunov
exponents on the underlying system is subtle, and not well understood. See
Ruelle [29] and also Bourgain, Jitomirskaya [9], [10] for a discussion and further references. We also mention the following simple application of the result
just stated, in the context of quasi-periodic Schrödinger cocycles:
Example 2. Let f : S 1 → S 1 be an irrational rotation. Given E ∈ R and
a continuous function V : S 1 → R, let A : S 1 → SL(2, R) be given by
E − V (θ) −1
A(θ) =
.
1
0
Then the cocycle determined by (A, f ) is a point of discontinuity for the
Lyapunov exponents, as functions of V ∈ C 0 (S 1 , R), if and only if the exponents are nonzero and E is in the spectrum of the associated Schrödinger operator. Compare [10]. This is because E is in the complement of the spectrum
if and only if the cocycle is uniformly hyperbolic, which for SL(2, R)-cocycles
is equivalent to domination.
We extend the two-dimensional result of Mañé–Bochi also in a different
direction, namely to symplectic diffeomorphisms on any compact symplectic
manifold. Firstly, we prove that continuity points for the Lyapunov exponents either are uniformly hyperbolic or have at least two Lyapunov exponents
equal to zero at almost every point. Consequently, generic symplectic C 1 diffeomorphisms either are Anosov or have vanishing Lyapunov exponents with
multiplicity at least 2 at almost every point.
Topological results in the vein of our present theorems were obtained
by Millionshchikov [22], in the early eighties, and by Bonatti, Dı́az, Pujals,
Ures [8], [12], in their recent characterization of robust transitivity for diffeomorphisms. A counterpart of the latter for symplectic maps was obtained
by Newhouse [25] in the seventies, and was recently extended by Arnaud [1].
Also recently, Dolgopyat, Pesin [13, §8] extended the perturbation technique
of [4] to one 4-dimensional case, as part of their construction of nonuniformly
hyperbolic diffeomorphisms on any compact manifold.
1.1. Dominated splittings. Let M be a compact manifold of dimension
d ≥ 2. Let f : M → M be a diffeomorphism and Γ ⊂ M be an f -invariant set.
Suppose for each x ∈ Γ one is given nonzero subspaces Ex1 and Ex2 such that
Tx M = Ex1 ⊕Ex2 , the dimensions of Ex1 and Ex2 are constant, and the subspaces
are Df -invariant: Dfx (Exi ) = Efi (x) for all x ∈ Γ and i = 1, 2.
Definition 1.1. Given m ∈ N, we say that TΓ M = E 1 ⊕ E 2 is an
m-dominated splitting if for every x ∈ Γ,
(1.1)
Dfxm |Ex2 · (Dfxm |Ex1 )−1 ≤
1
2
.
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JAIRO BOCHI AND MARCELO VIANA
We call TΓ M = E 1 ⊕ E 2 a dominated splitting if it is m-dominated for some
m ∈ N. Then we write E 1 E 2 .
Condition (1.1) means that, for typical tangent vectors, their forward
iterates converge to E 1 and their backward iterates converge to E 2 , at uniform
exponential rates. Thus, E 1 acts as a global hyperbolic attractor, and E 2
acts as a global hyperbolic repeller, for the dynamics induced by Df on the
projective bundle.
More generally, we say that a splitting TΓ M = E 1 ⊕ · · · ⊕ E k , into any
number of sub-bundles, is dominated if
E 1 ⊕ · · · ⊕ E j E j+1 ⊕ · · · ⊕ E k
for every 1 ≤ j < k.
We say that a splitting TΓ M = E 1 ⊕· · ·⊕E k , is dominated at x, for some point
x ∈ Γ, if it is dominated when restricted to the orbit {f n (x); n ∈ Z} of x.
1.2. Dichotomy for volume-preserving diffeomorphisms. Let µ be the
measure induced by some volume form. We indicate by Diff 1µ (M ) the set of
all µ-preserving C 1 diffeomorphisms of M , endowed with the C 1 topology. Let
f ∈ Diff 1µ (M ). By the theorem of Oseledets [26], for µ-almost every point
x ∈ M , there exist k(x) ∈ N, real numbers λ̂1 (f, x) > · · · > λ̂k(x) (f, x), and
k(x)
a splitting Tx M = Ex1 ⊕ · · · ⊕ Ex
of the tangent space at x, all depending
measurably on the point x, such that
(1.2)
lim
n→±∞
1
log Dfxn (v) = λ̂j (f, x)
n
for all v ∈ Exj {0}.
The Lyapunov exponents λ̂j (f, x) also correspond to the limits of 1/(2n) log ρn
as n → ∞ , where ρn represents the eigenvalues of Df n (x)∗ Df n (x). Let
λ1 (f, x) ≥ λ2 (f, x) ≥ · · · ≥ λd (f, x) be the Lyapunov exponents in nonincreasing order and each repeated with multiplicity dim Exj . Note that λ1 (f, x) +
· · · + λd (f, x) = 0, because f preserves volume. We say that the Oseledets
splitting is trivial at x when k(x) = 1, that is, when all Lyapunov exponents
vanish.
It should be stressed that these are purely asymptotic statements: the
limits in (1.2) are far from being uniform, in general. However, our first main
result states that for generic volume-preserving diffeomorphisms one does have
a lot of uniformity, over every orbit in a full measure subset:
Theorem 1. There exists a residual set R ⊂ Diff 1µ (M ) such that, for each
f ∈ R and µ-almost every x ∈ M , the Oseledets splitting of f is either trivial
or dominated at x.
For f ∈ R the ambient manifold M splits, up to zero measure, into disjoint invariant sets Z and D corresponding to trivial splitting and dominated
LYAPUNOV EXPONENTS
1427
splitting, respectively. Moreover, D may be written as an increasing union
D = ∪m∈N Dm of compact f -invariant sets, each admitting a dominated splitting of the tangent bundle.
If f ∈ R is ergodic then either µ(Z) = 1 or there is m ∈ N such that
µ(Dm ) = 1. The first case means that all the Lyapunov exponents vanish
almost everywhere. In the second case, the Oseledets splitting extends continuously to a dominated splitting of the tangent bundle over the whole ambient
manifold M .
Example 3. Let ft : N → N , t ∈ S 1 , be a smooth family of volumepreserving diffeomorphisms on some compact manifold N , such that ft = id for
t in some interval I ⊂ S 1 , and ft is partially hyperbolic for t in another interval
J ⊂ S 1 . Such families may be obtained, for instance, using the construction
of partially hyperbolic diffeomorphisms isotopic to the identity in [7]. Then
f : S 1 ×N → S 1 ×N , f (t, x) = (t, ft (x)) is a volume-preserving diffeomorphism
for which D ⊃ S 1 × J and Z ⊃ S 1 × I.
Thus, in general we may have 0 < µ(Z) < 1. However, we ignore whether
such examples can be made generic (see also Section 1.3).
Problem 1. Is there a residual subset of Diff 1µ (M ) for which invariant sets
with a dominated splitting have either zero or full measure?
Theorem 1 is a consequence of the following result about continuity of
Lyapunov exponents as functions of the dynamics. For j = 1, . . . , d − 1, define
LEj (f ) =
[λ1 (f, x) + · · · + λj (f, x)] dµ(x).
M
It is well-known that the functions f ∈ Diff 1µ (M ) → LEj (f ) are upper semicontinuous (see Proposition 2.2 below). Our next main theorem shows that
lower semi-continuity is much more delicate:
Theorem 2. Let f0 ∈ Diff 1µ (M ) be such that the map
f ∈ Diff 1µ (M ) → LE1 (f ), . . . , LEd−1 (f ) ∈ Rd−1
is continuous at f = f0 . Then for µ-almost every x ∈ M , the Oseledets splitting
of f0 is either dominated or trivial at x.
The set of continuity points of a semi-continuous function on a Baire
space is always a residual subset of the space (see e.g. [18, §31.X]); therefore
Theorem 1 is an immediate corollary of Theorem 2.
Problem 2. Is the necessary condition in Theorem 2 also sufficient for
continuity?
1428
JAIRO BOCHI AND MARCELO VIANA
Diffeomorphisms with all Lyapunov exponents equal to zero almost everywhere, or else whose Oseledets splitting extends to a dominated splitting
over the whole manifold, are always continuity points. Moreover, the answer
is affirmative in the context of linear cocycles, as we shall see.
1.3. Dichotomy for symplectic diffeomorphisms. Now we turn ourselves to
symplectic systems. Let (M 2q , ω) be a compact symplectic manifold without
boundary. We denote by µ the volume measure associated to the volume form
ω q = ω ∧ · · · ∧ ω. The space Sympl1ω (M ) of all C 1 symplectic diffeomorphisms
is a subspace of Diff 1µ (M ). We also fix a smooth Riemannian metric on M , the
particular choice being irrelevant for all purposes.
The Lyapunov exponents of symplectic diffeomorphisms have a symmetry
property: λj (f, x) = −λ2q−j+1 (f, x) for all 1 ≤ j ≤ q. (That is because in this
case the linear operator Df n (x)∗ Df n (x) is symplectic and so (see Arnold [3])
its spectrum is symmetric; the inverse of every eigenvalue is also an eigenvalue,
with the same multiplicity.) In particular, λq (x) ≥ 0 and LEq (f ) is the integral
of the sum of all nonnegative exponents. Consider the splitting
Tx M = Ex+ ⊕ Ex0 ⊕ Ex− ,
where Ex+ , Ex0 , and Ex− are the sums of all Oseledets spaces associated to
positive, zero, and negative Lyapunov exponents, respectively. Then dim Ex+ =
dim Ex− and dim Ex0 is even.
Theorem 3. Let f0 ∈ Sympl1ω (M ) be such that the map
f ∈ Sympl1ω (M ) → LEq (f ) ∈ R
is continuous at f = f0 . Then for µ-almost every x ∈ M , either dim Ex0 ≥ 2
or the splitting Tx M = Ex+ ⊕ Ex− is hyperbolic along the orbit of x.
In the second alternative, what we actually prove is that the splitting is
dominated at x. This is enough because, as we shall prove in Lemma 2.4,
for symplectic diffeomorphisms dominated splittings into two subspaces of the
same dimension are uniformly hyperbolic.
As in the volume-preserving case, the function f → LEq (f ) is continuous
on a residual subset R1 of Sympl1ω (M ). Also, we show that there is a residual
subset R2 ⊂ Sympl1ω (M ) such that for every f ∈ R2 either f is an Anosov
diffeomorphism or all its hyperbolic sets have zero measure. Taking R =
R1 ∩ R2 , we obtain:
Theorem 4. There exists a residual set R ⊂ Sympl1ω (M ) such that every
f ∈ R either is Anosov or has at least two zero Lyapunov exponents at almost
every point.
For d = 2 one recovers the two-dimensional result of Mañé–Bochi.
LYAPUNOV EXPONENTS
1429
1.4. Linear cocycles. Now we comment on corresponding statements for
linear cocycles. Let M be a compact Hausdorff space, µ a Borel regular probability measure, and f : M → M a homeomorphism that preserves µ. Given
a continuous map A : M → GL(d, R), one associates the linear cocycle
(1.3)
FA : M × Rd → M × Rd ,
FA (x, v) = (f (x), A(x)v).
Oseledets’ theorem extends to this setting, and so does the concept of dominated splitting; see Sections 2.1 and 2.2.
One is often interested in classes of maps A whose values have some specific form, e.g., belong to some subgroup G ⊂ GL(d, R). To state our results
in greater generality, we consider the space C(M, S) of all continuous maps
M → S, where S ⊂ GL(d, R) is a fixed set. We endow the space C(M, S)
with the C 0 -topology. We shall deal with sets S that satisfy an accessibility
condition:
Definition 1.2. Let S ⊂ GL(d, R) be an embedded submanifold (with
or without boundary). We call S accessible if for all C0 > 0 and ε > 0,
there are ν ∈ N and α > 0 with the following properties: Given ξ, η in
the projective space RPd−1 with (ξ, η) < α, and A0 , . . . , Aν−1 ∈ S with
A±1
i ≤ C0 , there exist A0 , . . . , Aν−1 ∈ S such that Ai − Ai < ε and
0 (ξ) = Aν−1 . . . A0 (η).
ν−1 . . . A
A
Example 4. Let G be a closed subgroup GL(d, R) which acts transitively
in the projective space RPd−1 . Then S = G is accessible and, in fact, we
may always take ν = 1 in the definition. See Lemma 5.12. So the most
common matrix groups are accessible, e.g., GL(d, R), SL(d, R), Sp(2q, R), as
well as SL(d, C), GL(d, C) (which are isomorphic to subgroups of GL(2d, R)).
(Compact groups are not of interest in our context, because all Lyapunov
exponents vanish identically.)
Example 5. The set of matrices of the type already mentioned in Example 2:
t −1
S=
; t ∈ R ⊂ GL(2, R)
1 0
is accessible. To see this, let ν = 2. If ξ and η are not too close to R(0, 1),
0 of A0 such that A
0 (ξ) = A0 (η), and
then we may find a small perturbation A
let A1 = A1 . In the other case, A0 (ξ) and A0 (η) must be close to R(1, 0); then
0 = A0 and find a suitable A
1 .
we take A
Theorem 5. Let S ⊂ GL(d, R) be an accessible set. Then A0 ∈ C(M, S)
is a point of continuity of
C(M, S) A → (LE1 (A), . . . , LEd−1 (A)) ∈ Rd−1
1430
JAIRO BOCHI AND MARCELO VIANA
if and only if the Oseledets splitting of the cocycle FA at x is either dominated
or trivial at µ-almost every x ∈ M .
Consequently, there exists a residual subset R ⊂ C(M, S) such that for
every A ∈ R and at almost every x ∈ X, either all Lyapunov exponents of FA
are equal or the Oseledets splitting of FA is dominated.
Corollary 1. Assume (f, µ) is ergodic. For any accessible set S ⊂
GL(d, R), there exists a residual subset R ⊂ C(M, S) such that every A ∈ R
either has all exponents equal at almost every point, or there exists a dominated splitting of M × Rd which coincides with the Oseledets splitting almost
everywhere.
Theorem 5 and the corollary remain true if one replaces C(M, S) by
L∞ (M, S). We only need f to be an invertible measure-preserving transformation.
It is interesting that an accessibility condition of control-theoretic type
was used by Nerurkar [24] to get nonzero exponents.
1.5. Extensions, related problems, and outline of the proof. Most of the
results stated above were announced in [5]. Actually, our Theorems 3 and 4
do not give the full strength of Theorem 4 in [5]. The difficulty is that the
symplectic analogue of our construction of realizable sequences is less satisfactory, unless the subspaces involved have the same dimension; see Remark 5.2.
Thus, the following question remains open (see also Remark 2.5):
Problem 3. Is it true that the Oseledets splitting of generic symplectic C 1
diffeomorphisms is either trivial or partially hyperbolic at almost every point?
Problem 4. For generic smooth families Rp → Diff 1µ (M ), Sympl1ω (M ),
C(M, S) (i.e. smooth in the parameters), what can be said of the Lebesgue
measure of the subset of parameters corresponding to zero Lyapunov exponents?
Problem 5. What are the continuity points of Lyapunov exponents in
r
Diff 1+r
µ (M ) or C (M, S) for r > 0?
Problem 6. Is the generic volume-preserving C 1 diffeomorphism ergodic
or, at least, does it have only a finite number of ergodic components?
The first question in Problem 6 was posed to us by A. Katok and the
second one was suggested by the referee. The theorem of Oxtoby, Ulam [27]
states that generic volume-preserving homeomorphisms are ergodic.
Let us close this introduction with a brief outline of the proof of Theorem 2.
Theorems 3 and 5 follow from variations of these arguments, and the other
main results are fairly direct consequences.
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LYAPUNOV EXPONENTS
Suppose the Oseledets splitting is neither trivial nor dominated, over a
positive Lebesgue measure set of orbits: for some i and for arbitrarily large m
there exist iterates y for which
(1.4)
Df m |Eyi− (Df m |Eyi+ )−1 >
1
2
k(y)
where Eyi+ = Ey1 ⊕ · · · ⊕ Eyi and Eyi− = Eyi+1 ⊕ · · · ⊕ Ey . The basic strategy is
to take advantage of this fact to, by a small perturbation of the map, cause a
vector originally in Eyi+ to move to Ezi− , z = f m (y), thus “blending” different
expansion rates.
More precisely, given a perturbation size ε > 0 we take m sufficiently large
with respect to ε. Then, given x ∈ M , for n much bigger than m we choose
an iterate y = f (x), with ≈ n/2, as in (1.4). By composing Df with small
rotations near the first m iterates of y, we cause the orbit of some Dfx (v) ∈ Eyi+
to move to Ezi− . In this way we find an ε-perturbation g = f ◦ h preserving
the orbit segment {x, . . . , f n (x)} and such that Dgxs (v) ∈ E i+ during the
first ≈ n/2 iterates and Dgxs (v) ∈ E i− during the last n − − m ≈ n/2
iterates. We want to conclude that Dgxn lost some expansion if compared to
Dfxn . To this end we compare the pth exterior products of these linear maps,
with p = dim E i+ . While ∧p (Dfxn ) ≈ exp(n(λ1 + · · · + λp )) we see that
λp + λp+1 ,
∧p (Dgxn ) exp n λ1 + · · · + λp−1 +
2
where the Lyapunov exponents are computed at (f, x). Notice that λp+1 =
λ̂i+1 is strictly smaller than λp = λ̂i . This local procedure is then repeated for
a positive Lebesgue measure set of points x ∈ M . Using (see Proposition 2.2)
1
log ∧p (Dg n ) dµ
LEp (g) = inf
n n
and a Kakutani tower argument, we deduce that LEp drops under such arbitrarily small perturbations, contradicting continuity.
Let us also comment on the way the C 1 topology comes into the proof.
It is very important for our arguments that the various perturbations of the
diffeomorphism close to each f s (y) do not interfere with each other, nor with
the other iterates of x in the time interval {0, . . . , n}. The way we achieve
this is by rescaling the perturbation g = f ◦ h near each f s (y) if necessary, to
ensure its support is contained in a sufficiently small neighborhood of the point.
In local coordinates w for which f s (y) is the origin, rescaling corresponds to
replacing h(w) by rh(w/r) for some small r > 0. Observe that this does not
affect the value of the derivative at the origin nor the C 1 norm of the map,
but it tends to increase C r norms for r > 1.
This paper is organized as follows. In Section 2 we introduce several
preparatory notions and results. In Section 3 we state and prove the main
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JAIRO BOCHI AND MARCELO VIANA
perturbation tool, the directions exchange Proposition 3.1. We use this proposition to prove Theorem 2 in Section 4, where we also deduce Theorem 1.
Section 5 contains a symplectic version of Proposition 3.1. This is used in Section 6 to prove Theorem 3, from which we deduce Theorem 4. Similar ideas,
in an easier form, are used in Section 7 to get Theorem 5.
2. Preliminaries
2.1. Lyapunov exponents, Oseledets splittings. Let M be a compact
Hausdorff space and π : E → M be a continuous finite-dimensional vector
bundle endowed with a continuous Riemann structure. A cocycle over a homeomorphism f : M → M is a continuous transformation F : E → E such
that π ◦ F = f ◦ π and Fx : Ex → Ef (x) is a linear isomorphism on each fiber
Ex = π −1 (x). Notice that (1.3) corresponds to the case when the vector bundle
is trivial.
2.1.1. Oseledets’ theorem. Let µ be any f -invariant Borel probability
measure in M . The theorem of Oseledets [26] states that for µ-almost every
point x there exists a splitting
(2.1)
Ex = Ex1 ⊕ · · · ⊕ Exk(x) ,
and real numbers λ̂1 (x) > · · · > λ̂k(x) (x) such that Fx (Exj ) = Efj (x) and
1
log Fxn (v) = λ̂j (x)
n→±∞ n
lim
for v ∈ Exj {0} and j = 1, . . . , k(x). Moreover, if J1 and J2 are any disjoint
subsets of the set of indexes {1, . . . , k(x)}, then
1
(2.2)
lim
Efj n (x) ,
Efj n (x) = 0.
log j∈J1
j∈J2
n→±∞ n
Let λ1 (x) ≥ λ2 (x) ≥ · · · ≥ λd (x) be the numbers λ̂j (x), each repeated with
multiplicity dim Exj and written in nonincreasing order. When the dependence
on F matters, we write λi (F, x) = λi (x). In the case when F = Df , we write
λi (f, x) = λi (F, x) = λi (x).
2.1.2. Exterior products. Given a vector space V and a positive integer p,
p
th
let
∧
d (V ) be the p exterior power of V . This is a vector space of dimension
p , whose elements are called p-vectors. It is generated by the p-vectors of
the form v1 ∧ · · · ∧ vp with vj ∈ V , called the decomposable p-vectors. A linear
map L : V → W induces a linear map ∧p (L) : ∧p (V ) → ∧p (W ) such that
∧p (L)(v1 ∧ · · · ∧ vp ) = L(v1 ) ∧ · · · ∧ L(vp ).
LYAPUNOV EXPONENTS
1433
If V has an inner product, then we always endow ∧p (V ) with the inner product
such that v1 ∧ · · · ∧ vp equals the p-dimensional volume of the parallelepiped
spanned by v1 , . . . , vp . See [2, §3.2.3].
More generally, there is a vector bundle ∧p (E), with fibers ∧p (Ex ), associated to E, and there is a vector bundle automorphism ∧p (F ), associated to F .
If the vector bundle E is endowed with a continuous inner product, then ∧p (E)
also is. The Oseledets data of ∧p (F ) can be obtained from that of F , as shown
by the proposition below. For a proof, see [2, Th. 5.3.1].
Proposition
2.1. The Lyapunov exponents (with multiplicity) λ∧p
i (x),
d
1 ≤ i ≤ p , of the automorphism ∧p (F ) at a point x are the numbers
λi1 (x) + · · · + λip (x),
where 1 ≤ i1 < · · · < ip ≤ d.
Let {e1 (x), . . . , ed (x)} be a basis of Ex such that
ei (x) ∈ Ex
for dim Ex1 + · · · + dim Ex−1 < i ≤ dim Ex1 + · · · + dim Ex .
Then the Oseledets space Exj,∧p of ∧p (F ) corresponding to the Lyapunov exponent λ̂j (x) is the sub-space of ∧p (Ex ) generated by the p-vectors
ei1 ∧ · · · ∧ eip ,
with 1 ≤ i1 < · · · < ip ≤ d and λi1 (x) + · · · + λip (x) = λ̂j (x).
2.1.3. Semi-continuity of integrated exponents. Let us indicate Λp (F, x) =
λ1 (F, x)+· · ·+λp (F, x), for p = 1, . . . , d−1. We define the integrated Lyapunov
exponent
LEp (F ) =
Λp (F, x) dµ(x).
M
More generally, if Γ ⊂ M is a measurable f -invariant subset, we define
LEp (F, Γ) =
Λp (F, x) dµ(x).
Γ
By Proposition 2.1, Λp (F, x) = λ1 (∧p F, x) and so LEp (F, Γ) = LE1 (∧p (F ), Γ).
When F = Df , we write Λp (f, x) = Λi (F, x) and LEp (f, Γ) = LEp (F, Γ).
Proposition 2.2. If Γ ⊂ M is a measurable f -invariant subset then
1
LEp (F, Γ) = inf
log ∧p (Fxn ) dµ(x).
n≥1 n Γ
Proof. The sequence an = Γ log ∧p (Fxn ) dµ is sub-additive (an+m ≤
an + am ); therefore lim ann = inf ann .
As a consequence of Proposition 2.2, the map f ∈ Diff 1µ (M ) → LEp (f ) is
upper semi-continuous, as mentioned in the introduction.
1434
JAIRO BOCHI AND MARCELO VIANA
2.2. Dominated splittings. Let Γ ⊂ M be an f -invariant set. A splitting
EΓ = E 1 ⊕ E 2 is dominated for F if it is F -invariant, the dimensions of Exi are
constant on Γ, and there exists m ∈ N such that, for every x ∈ Γ,
(2.3)
Fxm |Ex2 1
≤ .
m
1
m(Fx |Ex )
2
We denote m(L) = L−1 −1 the co-norm of a linear isomorphism L. The
dimension of the space E 1 is called the index of the splitting.
A few elementary properties of dominated decompositions follow. The
proofs are left to the reader.
Transversality. If EΓ = E 1 ⊕ E 2 is a dominated splitting then the angle
(Ex1 , Ex2 ) is bounded away from zero, over all x ∈ Γ.
Uniqueness. If EΓ = E 1 ⊕ E 2 and EΓ = Ê 1 ⊕ Ê 2 are dominated decompositions with dim E i = dim Ê i then E i = Ê i for i = 1, 2.
Continuity.
A dominated splitting EΓ = E 1 ⊕ E 2 is continuous, and
extends continuously to a dominated splitting over the closure of Γ.
2.3. Dominance and hyperbolicity for symplectic maps. We just recall
a few basic notions that are needed in this context, referring the reader to
Arnold [3] for definitions and fundamental properties of symplectic forms, manifolds, and maps.
Let (V, ω) be a symplectic vector space of dimension 2q. Given a subspace
W ⊂ V , its symplectic orthogonal is the space (of dimension 2q − dim W )
W ω = {w ∈ W ; ω(v, w) = 0 for all v ∈ V }.
The subspace W is called symplectic if W ω ∩ W = {0}; that is, ω|W ×W is a
nondegenerate form. W is called isotropic if W ⊂ W ω , that is, ω|W ×W ≡ 0.
The subspace W is called Lagrangian if W = W ω ; that is, it is isotropic and
dim W = q.
Now let (M, ω) be a symplectic manifold of dimension d = 2q. We also fix
in M a Riemannian structure. For each x ∈ M , let Jx : Tx M → Tx M be the
anti-symmetric isomorphism defined by ω(v, w) = Jx v, w for all v, w ∈ Tx M .
Denote
(2.4)
Cω = sup Jx±1 .
x∈M
In particular,
(2.5)
|ω(v, w)| ≤ Cω v w
for all v, w ∈ Tx M .
Lemma 2.3. If E, F ⊂ Tx M are two Lagrangian subspaces with E ∩ F =
{0} and α = (E, F ) then:
LYAPUNOV EXPONENTS
1435
(1) For every v ∈ E {0} there exists w ∈ F {0} such that
|ω(v, w)| ≥ Cω−1 sin α v w.
(2) If S : Tx M → Ty M is any symplectic linear map and β = (S(E), S(F ))
then
Cω−2 sin α ≤ m(S|E ) S|F ≤ Cω2 (sin β)−1 .
Proof. To prove part 1, let p : Tx M → F be the projection parallel
to E. Given a nonzero v ∈ E, take w = p(Jx v). Since E is isotropic,
ω(v, w) = ω(v, Jx v) = Jx v2 ≥ Cω−1 v Jx v. Also w ≤ p Jx v and
p = 1/ sin α, so that the claim follows.
To prove part 2, take a nonzero v ∈ E such that Sv/v = m(S|E ) and
let w be as in part 1. Then
Cω−1 sin α v w ≤ |ω(v, w)| = |ω(Sv, Sw)| ≤ Cω Sv Sw.
Thus m(S|E ) Sw/w ≥ Cω−2 sin α, proving the lower inequality in part 2.
The upper inequality follows from the lower one applied to S(F ), S(E) and
S −1 in the place of E, F , and S, respectively.
Lemma 2.4. Let f ∈ Sympl1ω (M ), and let x be a regular point. Assume
that λq (f, x) > 0, that is, there are no zero exponents. Let Ex+ and Ex− be the
sum of all Oseledets subspaces associated to positive and to negative Lyapunov
exponents, respectively. Then
(1) The subspaces Ex+ and Ex− are Lagrangian.
(2) If the splitting E + ⊕E − is dominated at x then E + is uniformly expanding
and E − is uniformly contracting along the orbit of x.
Proof. To prove part 1, we only have to show that the spaces Ex+ and Ex−
are isotropic. Take vectors v1 , v2 ∈ Ex− . Take ε > 0 with ε < λq (f, x). For
every large n and i = 1, 2, we have Dfxn vi ≤ e−nε vi . Hence, by (2.5),
|ω(v1 , v2 )| = |ω(Dfxn v1 , Dfxn v2 )| ≤ Cω e−2nε v1 v2 ,
that is, ω(v1 , v2 ) = 0. A similar argument, iterating backward, gives that Ex+
is isotropic.
Now assume that E + E − at x. Let α > 0 be a lower bound for
+
(E , E − ) along the orbit of x, and let C = Cω2 (sin α)−1 . By domination,
there exists m ∈ N such that
Dffmn (x) |E − m(Dffmn (x) |E + )
<
1
,
4C
for all n ∈ Z.
1436
JAIRO BOCHI AND MARCELO VIANA
By part 2 of Lemma 2.4, we have C −1 ≤ m(Dffmn (x) |E + ) Dffmn (x) |E − ≤ C.
Therefore
m(Dffmn (x) |E + ) > 2
and Dffmn (x) |E − <
1
2
for all n ∈ Z.
This proves part 2.
Remark 2.5. More generally, existence of a dominated splitting implies
partial hyperbolicity: If E ⊕ F is a dominated splitting, with dim E ≤ dim F,
then F splits invariantly as F = C ⊕ F , with dim F = dim E. Moreover,
the splitting E ⊕ C ⊕ F is dominated , E is uniformly expanding, and F is
uniformly contracting. This fact was pointed out by Mañé in [20]. Proofs
appeared recently in Arnaud [1], for dimension 4, and in [6], for arbitrary
dimension.
2.4. Angle estimation tools. Here we collect a few useful facts from elementary linear algebra. We begin by noting that, given any one-dimensional
subspaces A, B, and C of Rd , then
sin (A, B) sin (A + B, C) = sin (C, A) sin (C + A, B)
= sin (B, C) sin (B + C, A).
Indeed, this quantity is the 3-dimensional volume of the parallelepiped with
unit edges in the directions A, B and C. As a corollary, we get:
Lemma 2.6. Let A, B and C be subspaces (of any dimension) of Rd .
Then
sin (A, B + C) ≥ sin (A, B) sin (A + B, C).
Let v, w be nonzero vectors. For any α ∈ R, v + αw ≥ v sin (v, w),
with equality when α = v, w/w2 . Given L ∈ GL(d, R), let β = Lv, Lw/
Lw2 and z = v + βw. By the previous remark, z ≥ v sin (v, w) and
Lz = Lv sin (Lv, Lw). Therefore
(2.6)
sin (Lv, Lw) =
Lz
m(L)v
≥
sin (v, w).
Lv
Lv
As a consequence of (2.6), we have:
Lemma 2.7. Let L : Rd → Rd be a linear map and let v, w be nonzero
vectors. Then
m(L)
sin (Lv, Lw)
L
≤
≤
.
L
sin (v, w)
m(L)
Thus L/m(L) measures how much angles can be distorted by L. At
last, we give a bound for this quantity when d = 2.
LYAPUNOV EXPONENTS
1437
Lemma 2.8. Let L : R2 → R2 be an invertible linear map and let
v, w ∈ R2 be linearly independent unit vectors. Then
Lv Lw
1
1
L
≤ 4 max
,
.
m(L)
Lw Lv sin (v, w) sin (Lv, Lw)
Proof. We may assume that L is not conformal, for in the conformal case
the left-hand side is 1 and the inequality is obvious. Let Rs be the direction
most contracted by L, and let θ, φ ∈ [0, π] be the angles that the directions
Rv and Rw, respectively, make with Rs. Suppose that Lv ≥ Lw. Then
φ ≤ θ and so (v, w) ≤ 2θ. Hence
Lv ≥ L sin θ ≥ 12 L sin 2θ ≥ 12 L sin (v, w).
Moreover, |det L| = m(L)L and
Lv Lw sin (Lv, Lw) = |det L| sin (v, w).
The claim is an easy consequence of these relations.
2.5. Coordinates, metrics, neighborhoods. Let (M, ω) be a symplectic
manifold of dimension d = 2q ≥ 2. According to Darboux’s theorem, there
exists an atlas A∗ = {ϕi : Vi∗ → Rd } of canonical local coordinates, that is,
such that
(ϕi )∗ ω = dx1 ∧ dx2 + · · · + dx2q−1 ∧ dx2q
for all i. Similarly, cf. [23, Lemma 2], given any volume structure β on a
d-dimensional manifold M , one can find an atlas A∗ = {ϕi : Vi∗ → Rd }
consisting of charts ϕi such that
(ϕi )∗ β = dx1 ∧ · · · ∧ dxd .
In either case, assuming M is compact one may choose A∗ finite. Moreover, we may always choose A∗ so that every Vi∗ contains the closure of an
open set Vi , such that the restrictions ϕi : Vi → Rd still form an atlas of M .
The latter will be denoted A. Let A∗ and A be fixed once and for all.
By compactness, there exists r0 > 0 such that for each x ∈ M , there exists
i(x) such that the Riemannian ball of radius r0 around x is contained in Vi(x) .
For definiteness, we choose i(x) smallest with this property. For technical
convenience, when dealing with the point x we express our estimates in terms
of the Riemannian metric · = ·x defined on that ball of radius r0 by v =
Dϕi(x) v. Observe that these Riemannian metrics are (uniformly) equivalent
to the original one on M , and so there is no inconvenience in replacing one by
the other.
We may also view any linear map A : Tx1 M → Tx2 M as acting on Rd ,
using local charts ϕi(x1 ) and ϕi(x2 ) . This permits us to speak of the distance
1438
JAIRO BOCHI AND MARCELO VIANA
A − B between A and another linear map B : Tx3 M → Tx4 M whose base
points are different:
A − B = D2 AD1−1 − D4 BD3−1 ,
where Dj = (Dϕi(xj ) )xj .
For x ∈ M and r > 0 small (relative to r0 ), Br (x) will denote the ball
of radius
r around x relative to the new metric. In other words, Br (x) =
ϕ−1
B(ϕ
i(x) (x), r) . We assume that r is small enough so that the closure of
i(x)
∗ .
Br (x) is contained in Vi(x)
Definition 2.9. Let ε0 > 0. The ε0 -basic neighborhood U(id, ε0 ) of the
identity in Diff 1µ (M ), or in Sympl1ω (M ), is the set U(id, ε0 ) of all h ∈ Diff 1µ (M ),
or h ∈ Sympl1ω (M ), such that h±1 (V i ) ⊂ Vi∗ for each i and
h(x) ∈ Bε0 (x)
and Dhx − I < ε0
for every x ∈ M .
For a general f ∈ Diff 1µ (M ), or f ∈ Sympl1ω (M ), the ε0 -basic neighborhood
U(f, ε0 ) is defined by: g ∈ U(f, ε0 ) if and only if f −1 ◦ g ∈ U(id, ε0 ) or g ◦ f −1 ∈
U(id, ε0 ).
2.6. Realizable sequences. The following notion, introduced in [4], is crucial to the proofs of Theorems 1 through 4. It captures the idea of sequence
of linear transformations that can be (almost) realized on subsets with large
relative measure as tangent maps of diffeomorphisms close to the original one.
Definition 2.10. Given f ∈ Diff 1µ (M ) or f ∈ Sympl1ω (M ), constants
ε0 > 0, and 0 < κ < 1, and a nonperiodic point x ∈ M , we call a sequence of
linear maps (volume-preserving or symplectic)
L
L
Ln−1
0
1
Tf x M −→
. . . −−−→ Tf n x M
Tx M −→
an (ε0 , κ)-realizable sequence of length n at x if the following holds:
For every γ > 0 there is r > 0 such that the iterates f j (B r (x)) are twoby-two disjoint for 0 ≤ j ≤ n, and given any nonempty open set U ⊂ Br (x),
there are g ∈ U(f, ε0 ) and a measurable set K ⊂ U such that
j
(i) g equals f outside the disjoint union n−1
j=0 f (U );
(ii) µ(K) > (1 − κ)µ(U );
(iii) if y ∈ K then Dggj y − Lj < γ for every 0 ≤ j ≤ n − 1.
Some basic properties of realizable sequences are collected in the following:
Lemma 2.11. Let f ∈ Diff 1µ (M ) or f ∈ Sympl1ω (M ), x ∈ M not periodic
and n ∈ N.
(1) The sequence {Dfx , . . . , Dff n−1 (x) } is (ε0 , κ)-realizable for every ε0 and
κ (called a trivial realizable sequence).
LYAPUNOV EXPONENTS
1439
(2) Let κ1 , κ2 ∈ (0, 1) be such that κ = κ1 + κ2 < 1. If {L0 , . . . , Ln−1 }
is (ε0 , κ1 )-realizable at x, and {Ln , . . . , Ln+m−1 } is (ε0 , κ2 )-realizable at
f n (x), then {L0 , . . . , Ln+m−1 } is (ε0 , κ)-realizable at x.
−1
(3) If {L0 , . . . , Ln−1 } is (ε0 , κ)-realizable at x, then {L−1
n−1 , . . . , L0 } is an
n
−1
(ε0 , κ)-realizable sequence at f (x) for the diffeomorphism f .
Proof. The first claim is obvious. For the second one, fix γ > 0. Let
r1 be the radius associated to the (ε0 , κ1 )-realizable sequence, and r2 be the
radius associated to the (ε0 , κ2 )-realizable sequence. Fix 0 < r < r1 such that
f n (Br (x)) ⊂ B(f n (x), r2 ). Then the f j (B r (x)) are two-by-two disjoint for
0 ≤ j ≤ n + m. Given an open set U ⊂ Br (x), the realizability of the first
sequence gives us a diffeomorphism g1 ∈ U(f, ε0 ) and a measurable set K1 ⊂ U .
Analogously, for the open set f n (U ) ⊂ B(f n (x), r2 ) we find g2 ∈ U(f, ε0 ) and a
measurable set K2 ⊂ f n (U ). Then define a diffeomorphism g as g = g1 inside
U ∪ · · · ∪ f n−1 (U ) and g = g2 inside f n (U ) ∪ · · · ∪ f n+m−1 (U ), with g = f
elsewhere. Consider also K = K1 ∩ g −n (K2 ). Using the fact that g preserves
volume, one checks that g and K satisfy the conditions in Definition 2.10. For
claim 3, notice that U(f, ε0 ) = U(f −1 , ε0 ).
The next lemma makes it simpler to verify that a sequence is realizable:
we only have to check the conditions for certain open sets U ⊂ Br (x).
Definition 2.12. A family of open sets {Wα } in Rd is a Vitali covering of
W = ∪α Wα if there is C > 1 and for every y ∈ W , there are sequences of sets
Wαn y and positive numbers sn → 0 such that
Bsn (y) ⊂ Wαn ⊂ BCsn (y)
for all n ∈ N.
A family of subsets {Uα } of M is a Vitali covering of U = ∪α Uα if each Uα
is contained in the domain of some chart ϕi(α) in the atlas A, and the images
{ϕi(α) (Uα )} form a Vitali covering of W = ϕ(U ), in the previous sense.
Lemma 2.13. Let f ∈ Diff 1µ (M ) or f ∈ Sympl1ω (M ), and set ε0 > 0 and
κ > 0. Consider any sequence Lj : Tf j (x) M → Tf j+1 (x) M , 0 ≤ j ≤ n − 1 of
linear maps at a nonperiodic point x, and let ϕ : V → Rd be a chart in the
atlas A, with V x. Assume the conditions in Definition 2.10 are valid for
every element of some Vitali covering {Uα } of Br (x). Then the sequence Lj is
(ε0 , κ)-realizable.
Proof. Let U be an arbitrary open subset of Br (x). By Vitali’s covering
lemma (see [21]), there is a countable family of two-by-two disjoint sets Uα
covering U up to a zero Lebesgue measure subset. Thus we can find a finite
family of Uα with disjoint closures such that µ (U − α Uα ) is as small as we
please. For each Uα there are, by hypothesis, a perturbation gα ∈ U(f, ε0 ) and
1440
JAIRO BOCHI AND MARCELO VIANA
a measurable set Kα ⊂ Uα with the properties (i)–(iii) of Definition 2.10. Let
K = Kα and define g as being equal to gα on each f j (Uα ) with 0 ≤ j ≤
n − 1. Then g ∈ U(f, ε0 ) and the pair (g, K) have the properties required by
Definition 2.10.
3. Geometric consequences of nondominance
The aim of this section is to prove the following key result, from which we
shall deduce Theorem 2 in Section 4:
Proposition 3.1. When f ∈ Diff 1µ (M ), ε0 > 0 and 0 < κ < 1, if m ∈ N
is sufficiently large then the following holds: Let y ∈ M be a nonperiodic point
and assume that there is a nontrivial splitting Ty M = E ⊕ F such that
Dfym |F 1
≥ .
m
m(Dfy |E )
2
Then there exists an (ε0 , κ)-realizable sequence {L0 , . . . , Lm−1 } at y of length
m and there are nonzero vectors v ∈ E and w ∈ Dfym (F ) such that
Lm−1 . . . L0 (v) = w.
3.1. Nested rotations. Here we present some tools for the construction of
realizable sequences. The first one yields sequences of length 1:
Lemma 3.2. Given f ∈ Diff 1µ (M ), ε0 > 0, κ > 0, there exists ε > 0 with
the following properties:
Suppose there are a nonperiodic point x ∈ M , a splitting Tx M = X ⊕ Y
: Y → Y with
with X ⊥ Y and dim Y = 2, and an elliptic linear map R
R − I < ε. Consider the linear map R : Tx M → Tx M given by R(u + v) =
u + R(v),
for u ∈ X, v ∈ Y . Then {Dfx R} is an (ε0 , κ)-realizable sequence of
length 1 at x and {R Dff −1 (x) } is an (ε0 , κ)-realizable sequence of length 1 at
the point f −1 (x).
We call a linear isomorphism of a 2-dimensional space elliptic if its eigenvalues are not real; this means the map is a rotation, relative to some basis of
the space.
We also need to construct long realizable sequences. Part 2 of Lemma 2.11
provides a way to do this, by concatenation of shorter sequences. However,
simple concatenation is far too crude for our purposes because it worsens κ;
the relative measure of the set where the sequence can be (almost) realized decreases when the sequence increases. This problem is overcome by Lemma 3.3
below, which allows us to obtain certain nontrivial realizable sequences with
arbitrary length while keeping κ controlled.
1441
LYAPUNOV EXPONENTS
In short terms, we do concatenate several length 1 sequences, of the type
given by Lemma 3.2, but we also require that the supports of successive perturbations be mapped one to the other. More precisely, there is a domain
C0 ⊂ Tx M invariant under the sequence, in the sense that Lj−1 . . . L0 (C0 ) =
Dfxj (C0 ) for all j. Following [4], where a similar notion was introduced for
the 2-dimensional setting, we call such Lj nested rotations. When d > 2 the
domain C0 is not compact; indeed it is the product C0 = X0 ⊕ B0 of a codimension 2 subspace X0 by an ellipse B0 ⊂ X0⊥ .
Let us fix some terminology to be used in the sequel. If E is a vector space
with an inner product and F is a subspace of E, we endow the quotient space
E/F with the inner product that makes v ∈ F ⊥ → (v +F ) ∈ E/F an isometry.
If E is another vector space, any linear map L : E → E induces a linear map
L/F : E/F → E /F , where F = L(F ). If E has an inner product, then we
indicate by L/F the usual operator norm.
Lemma 3.3. When f ∈ Diff 1µ (M ), ε0 > 0, κ > 0, there exists ε > 0 with
the following properties: Suppose there are a nonperiodic point x ∈ M , an
integer n ≥ 1, and, for j = 0, 1, . . . , n − 1,
• codimension 2 spaces Xj ⊂ Tf j (x) M such that Xj = Dfxj (X0 );
• ellipses Bj ⊂ (Tf j (x) M )/Xj centered at zero with Bj = (Dfxj /X0 )(B0 );
j : (Tf j (x) M )/Xj → (Tf j (x) M )/Xj such that R
j (Bj ) ⊂ Bj
• linear maps R
j − I < ε.
and R
Consider the linear maps Rj : Tf j (x) M → Tf j (x) M such that Rj restricted to
j . Define
Xj is the identity, Rj (Xj⊥ ) = Xj⊥ and Rj /Xj = R
Lj = Dff j (x) Rj : Tf j (x) M → Tf j+1 (x) M
for 0 ≤ j ≤ n − 1.
Then {L0 , . . . , Ln−1 } is an (ε0 , κ)-realizable sequence of length n at x.
We shall prove Lemma 3.3 in Section 3.1.2. Notice that Lemma 3.2 is
contained in Lemma 3.3: take n = 1 and use also part 3 of Lemma 2.11.
Actually, Lemma 3.2 also follows from the forthcoming Lemma 3.4.
3.1.1. Cylinders and rotations. We call a cylinder any affine image C in Rd
of a product B d−i × B i , where B j denotes a ball in Rj . If ψ is the affine map,
the axis A = ψ(B d−i × {0}) and the base B = ψ({0} × B i ) are ellipsoids. We
also write C = A⊕B. The cylinder is called right if A and B are perpendicular.
The case we are most interested in is when i = 2.
The present section contains three preliminary lemmas that we use in the
proof of Lemma 3.3. The first one explains how to rotate a right cylinder,
while keeping the complement fixed. The assumption a > τ b means that the
1442
JAIRO BOCHI AND MARCELO VIANA
cylinder C is thin enough, and it is necessary for the C 1 estimate in part (ii)
of the conclusion.
Lemma 3.4. Given ε0 > 0 and 0 < σ < 1, there is ε > 0 with the
following properties: Suppose there are a splitting Rd = X ⊕ Y with X ⊥ Y
and dim Y = 2, a right cylinder A ⊕ B centered at the origin with A ⊂ X and
: Y → Y such that R(B)
− I < ε.
B ⊂ Y , and a linear map R
= B and R
Then there exists τ > 1 such that the following holds:
for
Let R : Rd → Rd be the linear map defined by R(u + v) = u + Rv,
u ∈ X, v ∈ Y . For a, b > 0 consider the cylinder C = aA ⊕ bB. If a > τ b and
diam C < ε0 then there is a C 1 volume-preserving diffeomorphism h : Rd → Rd
satisfying
(i) h(z) = z for every z ∈
/ C and h(z) = R(z) for every z ∈ σC;
(ii) h(z) − z < ε0 and Dhz − I < ε0 for all z ∈ Rd .
Proof. We choose ε > 0 small enough so that
(3.1)
18ε
< ε0 .
1−σ
R be as in the statement of the lemma. Let {e1 , . . . , ed }
Let A, B, X, Y , R,
be an orthonormal basis of Rd such that e1 , e2 ∈ Y are in the directions of the
axes of the ellipse B and ej ∈ X for j = 3, . . . , d. We shall identify vectors
v = xe1 + ye2 ∈ Y with the coordinates (x, y). Then there are constants λ ≥ 1
and ρ > 0 such that B = {(x, y); λ−2 x2 + λ2 y 2 ≤ ρ2 }. Relative to the basis
{e1 , e2 }, let
λ 0
cos α − sin α
and Rα =
Hλ =
.
0 λ−1
sin α cos α
= Hλ Rα H −1 for some α. Besides,
The assumption R(B)
= B implies that R
λ
− I < ε implies
the condition R
(3.2)
− I)(0, 1) < ε.
λ2 |sin α| ≤ (R
Let ϕ : R → R be a C ∞ function such that ϕ(t) = 1 for t ≤ σ, ϕ(t) = 0
for t ≥ 1, and 0 ≤ −ϕ (t) ≤ 2/(1 − σ) for all t. Define smooth maps ψ : Y → R
and g̃t : Y → Y by
ψ(x, y) = αϕ( x2 + y 2 ) and g̃t (x, y) = Rϕ(t)ψ(x,y) (x, y).
On the one hand, g̃t (x, y) = (x, y) if either t ≥ 1 or x2 + y 2 ≥ 1. On the other
hand, g̃t (x, y) = Rα (x, y) if t ≤ σ and x2 + y 2 ≥ σ 2 . We are going to check
that the derivative of g̃t is close to the identity if ε is close to zero; note that
1443
LYAPUNOV EXPONENTS
| sin α| is also close to zero, by (3.2). We have
cos(tψ) − sin(tψ)
−x sin(tψ) − y cos(tψ) D(g̃t )(x,y) =
+
· t∂x ψ t∂y ψ
sin(tψ) cos(tψ)
x cos(tψ) − y sin(tψ)
= Rtψ(x,y) + t Rπ/2+tψ(x,y) (x, y) · Dψ(x,y) .
Consider 0 ≤ t ≤ 1 and x2 + y 2 ≤ 1. Then
D(g̃t )(x,y) − I = Rtψ(x,y) − I + Rπ/2+tψ(x,y) (x, y) · Dψ(x,y) ≤ sin tψ(x, y) + 2αxϕ (x2 + y 2 ) , 2αyϕ (x2 + y 2 ) .
Taking ε small enough, we may suppose that α ≤ 2| sin α|. In view of the
choice of ϕ and ψ, this implies
(3.3)
D(g̃t )(x,y) − I ≤ |sin α| + 4|α|/(1 − σ) ≤ 9|sin α|/(1 − σ).
We also need to estimate the derivative with respect to t:
∂t g̃(x, y) ≤ ϕ (t)ψ(x, y)Rπ/2+tψ(x,y) (x, y) ≤ 4|sin α|/(1 − σ).
(3.4)
Now define gt : Y → Y by gt = Hλ ◦ g̃t ◦ Hλ−1 . Each gt is an area-preserving
diffeomorphism equal to the identity outside B. Thus
(3.5)
gt (x, y) − (x, y) < diam B,
= Hλ Rα H −1 on σB for all t ≤ σ.
for every (x, y) ∈ B. Moreover, gt = R
λ
By (3.3),
−1 9|sin
α|
2
,
D(gt )(x,y) − I = Hλ D(g̃t )(λ−1 x,λy) − I Hλ ≤ λ
1−σ
and, applying (3.2) and (3.1), we deduce that
(3.6)
D(gt )(x,y) − I <
9ε
ε0
<
1−σ
2
for all (x, y) ∈ B. Similarly, by (3.4),
(3.7)
∂t gt (x, y) ≤ λ ∂t g̃t (λ
2
−1
x, λy) ≤ λ
2
4|sin α|
1−σ
<
ε0
.
2
Now let Q : X → R be a quadratic form such that A = {u ∈ X; Q(u) ≤ 1},
and let q : Rd → X and p : Rd → Y be the orthogonal projections. Given
a, b > 0, define h : Rd → Rd by
h(z) = z + bga−2 Q(z ) (b−1 z ),
where z = q(z) and z = p(z).
It is clear that h is a volume-preserving diffeomorphism. The subscript t =
a−2 Q(z ) is designed so that t ≤ 1 if and only if z ∈ aA. Then h(z) = z if
) = R(z) if z ∈ σaA
either z ∈
/ aA or z ∈
/ bB. Moreover, h(z) = z + R(z
1444
JAIRO BOCHI AND MARCELO VIANA
and z ∈ σbB. This proves property (i) in the statement. The hypothesis
diam C < ε0 and (3.5) give
h(z) − z = bga−2 Q(z ) (b−1 z ) − b−1 z < b diam B ≤ diam(aA ⊕ bB) < ε0
which is the first half of (ii). Finally, fix τ > 1 such that DQu ≤ τ u for
all u ∈ Rd , and assume that a > τ b. Clearly,
Dh = q +
b
(∂t g)(DQ)q + (Dg)p.
a2
By (3.6), (3.7), and the fact that q = p = 1 (these are orthogonal projections),
b
(∂t g)(DQ)q + (Dg − I)p
a2
b
≤ 2 ∂t g τ a q + Dg − I p < ε0 .
a
Dh − I ≤ This completes the proof of property (ii) and the lemma.
The second of our auxiliary lemmas says that the image of a small cylinder
by a C 1 diffeomorphism h contains the image by Dh of a slightly shrunk
cylinder. Denote C(y, ρ) = ρC + y, for each y ∈ Rd and ρ > 0.
Lemma 3.5. Let h : Rd → Rd be a C 1 diffeomorphism with h(0) = 0,
C ⊂ Rd be a cylinder centered at 0, and 0 < λ < 1. Then there exists r > 0
such that for any C(y, ρ) ⊂ Br (0),
h(C(y, ρ)) ⊃ Dh0 (C(0, λρ)) + h(y).
Proof. Fix a norm ·0 in Rd for which C = {z ∈ Rd ; z0 < 1}. Such a
norm exists because C is convex and C = −C. Let H = Dh0 and g : Rd → Rd
be such that h = H ◦ g. Since g is C 1 and Dg0 = I, we have
g(z) − g(y) = z − y + ξ(z, y)
with
ξ(z, y)
= 0.
(z,y)→(0,0) z − y0
lim
Choose r > 0 such that z, y ≤ r ⇒ ξ(z, y)0 < (1 − λ)z − y0 (where
· denotes the Euclidean norm in Rd ). Now suppose C(y, ρ) ⊂ Br (0), and let
z ∈ ∂C(y, ρ). Then z − y0 = ρ and
g(z) − g(y)0 ≥ z − y0 − ξ(z, y)0 > λρ.
This proves that the sets g(∂C(y, ρ)) − g(y) and λC are disjoint. Applying the
linear map H, we find that h(∂C(y, ρ)) − h(y) and λHC are disjoint. From
topological arguments, h(C(y, ρ)) − h(y) ⊃ λHC.
1445
LYAPUNOV EXPONENTS
The third lemma says that a linear image of a sufficiently thin cylinder
contains a right cylinder with almost the same volume. The idea is shown in
Figure 1. The proof of the lemma is left to the reader.
Lemma 3.6. Let A⊕B be a cylinder centered at the origin, L : Rd → Rd be
a linear isomorphism, A1 = L(A) and B1 = p(L(B)), where p is the orthogonal
projection onto the orthogonal complement of A1 . Then, given any 0 < λ < 1,
there exists τ > 1 such that if a > τ b,
L aA ⊕ bB ⊃ λaA1 ⊕ bB1 .
bB1
L
bB
aA
bL(B)
λaA1
Figure 1: Truncating a thin cylinder to make it right
3.1.2. Proof of the nested rotations Lemma 3.3. Let f , ε0 , and κ be given.
Define σ = (1−κ)1/2d and then take ε > 0 as given by Lemma 3.4. Now let x, n,
j , Rj , Lj be as in the statement. We want to prove that {L0 , . . . , Ln }
Xj , Bj , R
is an (ε0 , κ)-realizable sequence of length n at x; cf. Definition 2.10.
In short terms, we use Lemma 3.4 to construct the realization g at each
iterate. The subset U K, where we have no control over the approximation,
has two sources: Lemma 3.4 gives h = R only on a slightly smaller cylinder σC;
and we need to straighten out (Lemma 3.5) and to “rightify” (Lemma 3.6)
our cylinders at each stage. These effects are made small by consideration of
cylinders that are small and very thin. That is how we get U K with relative
volume less than κ, independently of n.
For clearness we split the proof into three main steps:
Step 1. Fix any γ > 0. We explain how to find r > 0 as in Definition 2.10.
We consider local charts ϕj : Vj → Rd with ϕj = ϕi(f j x) and Vj = Vi(f j x) , as
introduced in Section 2.5. Let r > 0 be small enough so that
• f j (B r (x)) ⊂ Vj∗ for every j = 0, 1 . . . , n;
• the sets f j (B r (x)) are two-by-two disjoint;
• Dfz − Dff j (x) Rj < γ for every z ∈ f j (Br (x)) and j = 0, 1 . . . , n.
1446
JAIRO BOCHI AND MARCELO VIANA
We use local charts to translate the situation to Rd . Let fj = ϕj+1 ◦f ◦ϕ−1
j
be the expression of f in local coordinates near f j (x) and f j+1 (x). To simplify
the notation, we suppose that each ϕj has been composed with a translation
to ensure ϕj (f j (x)) = 0 for all j. Up to identification of tangent spaces via
the charts ϕj and ϕj+1 , we have Lj = (Dfj )0 Rj .
Let A0 ⊂ X0 be any ellipsoid centered at the origin (a ball, for example),
and let Aj = Dfxj (A0 ) for j ≥ 1. We identify (Tf j (x) M )/Xj with Xj⊥ , so that
we may consider Bj ⊂ Xj⊥ . In these terms, the assumption Bj = (Dfxj /X0 )(B0 )
means that Bj is the orthogonal projection of Dfxj (B0 ) onto Xj⊥ .
Fix 0 < λ < 1 close enough to 1 so that λ4n(d−1) > 1 − κ. Let τj > 1 be
associated to the data (Aj ⊕ Bj , (Dfj )0 , λ) by Lemma 3.6; if a > τj b then
(3.8)
(Dfj )0 (aAj ⊕ bBj ) ⊃ λaAj+1 ⊕ bBj+1 .
j ) by Lemma 3.4.
Let τj > 1 be associated to the data (ε0 , σ, Xj ⊕Xj⊥ , Aj ⊕Bj , R
Fix a0 > 0 and b0 > 0 such that
(3.9)
a0 > b0 λ−n max{τj , τj ; 0 ≤ j ≤ n − 1}.
For 0 ≤ j ≤ n, define Cj = λ2j a0 Aj ⊕ λj b0 Bj . For z ∈ Rd and ρ > 0, denote
Cj (z, ρ) = ρCj + z. Applying Lemma 3.5 to the data (fj , Cj , λ) we get rj > 0
such that
(3.10)
C(z, ρ) ⊂ Brj (0)
⇒
fj (Cj (z, ρ)) ⊃ (Dfj )0 (Cj (0, λρ)) + fj (z).
Now take r > 0 such that r < r and, for each j = 1, . . . , m − 1,
(3.11)
fj−1 . . . f0 (Br (0)) ⊂ Brj (0).
Step 2. Let U be fixed. We find g ∈ U(f, ε0 ) and K ⊂ U as in Definition 2.10. We take advantage of Lemma 2.13: it suffices to consider open sets
of the form U = ϕ−1
0 (C0 (y0 , ρ)), because the cylinders C0 (y0 , ρ) contained in
Br (0) constitute a Vitali covering.
We claim that, for each j = 0, 1, . . . , m − 1, and every t ∈ [0, ρ],
(3.12)
Cj (yj , t) ⊂ fj−1 . . . f0 (Br (0))
and
(3.13)
fj (Cj (yj , t)) ⊃ Cj+1 (yj+1 , t).
For j = 0, relation (3.12) means C0 (y0 , t) ⊂ Br (0), which is true by assumption.
We proceed by induction. Assume (3.12) holds for some j ≥ 0. Then, by (3.11)
and (3.10),
fj (Cj (yj , t)) ⊃ (Dfj )0 (Cj (0, λt)) + yj+1
= (Df0 )0 (λ2j+1 ta0 Aj ) ⊕ (λj+1 tb0 Bj ) + yj+1 .
LYAPUNOV EXPONENTS
1447
Relation (3.9) implies that λ2j+1 ta0 > τj (λj+1 tb0 ). So, we may use (3.8) to
conclude that
fj (Cj (yj , t)) ⊃ (λ2j+2 ta0 Aj ) ⊕ (λj+1 tb0 B0 ) + yj+1 = Cj+1 (yj+1 , t).
This proves that (3.13) holds for the same value of j. Moreover, it is clear that
if (3.13) holds for all 0 ≤ i ≤ j then (3.12) is true with j + 1 in the place of j.
This completes the proof of (3.12) and (3.13).
Condition (3.9) also implies λ2j a0 > τj (λj b0 ). So, we may use Lemma 3.4
(centered at yj ) to find a volume-preserving diffeomorphism hj : Rd → Rd such
that
(1) hj (z) = z for all z ∈
/ Cj (yj , ρ) and hj (z) = yj + Rj (z − yj ) for all z ∈
Cj (yj , σρ) and, consequently,
(3.14)
hj (Cj (yj , σρ)) = Cj (yj , σρ)
and hj (Cj (yj , ρ)) = Cj (yj , ρ).
(1) hj (z) − z < ε0 and (Dhj )z − I < ε0 for all z ∈ Rd .
Rj is the linear map Tf j (x) → Tf j+1 (x) in the statement of the theorem or,
more precisely, its expression in local coordinates ϕj . Let
Sj = ϕ−1
j ({z; h(z) = z}) ⊂ M.
By Property 1 above and the inclusion (3.12),
j
Sj ⊂ ϕ−1
j (fj−1 . . . f0 (Br (0))) = f (Br (x)).
In particular, the sets Sj have pairwise disjoint closures. This permits us to
define a diffeomorphism g ∈ Diff 1µ (M ) by
−1
ϕj+1 ◦ (fj ◦ hj ) ◦ ϕj on Sj for each 0 ≤ j ≤ n − 1
g=
f
outside S0 · · · Sn−1 .
Property 2 above gives that f −1 ◦ g ∈ U(id, ε0 ), and so g ∈ U(f, ε0 ).
Step 3.
Now we define K ⊂ U and check the conditions (i)–(iii) in
Definition 2.10. By construction, hj = id outside Cj (yj , ρ), and so
ϕ−1
j+1 ◦ (fj ◦ hj ) ◦ ϕj = f
outside ϕ−j (Cj (yj , ρ)).
Using (3.13) and (3.14), we have ϕ−j (Cj (yj , ρ)) ⊂ f j (U ) for all 0 ≤ j ≤ n − 1.
Recall that U = ϕ−1
0 (C0 (y0 , ρ)). Hence, g = j outside the disjoint union
j (U ). This proves condition (i).
n−1
f
j=0
Define K = g −n (ϕ−1
n (Cn (yn , σρ))). Using (3.13) and (3.14) in the same
way as before, we see that K ⊂ U . Also, since all the maps f , g, hj , ϕj are
volume-preserving, and all the cylinders Cj (yj , ρ), Cj (yj , σρ), are right,
vol K
vol(σρ λ2n aAn ⊕ σρ λn bBn )
(λ2n σ)d−2 vol An (λn σ)2 vol Bn
.
=
=
vol U
vol(ρaA0 ⊕ ρbB0 )
vol A0 vol B0
1448
JAIRO BOCHI AND MARCELO VIANA
Notice also that vol An vol Bn = vol A0 vol B0 , since the cylinders
Dfxn (A0 ⊕ B0 )
and An ⊕ Bn differ by a sheer. So, the right-hand side is equal to λ2n(d−1) σ d .
Now, this expression is larger than 1−κ, because we have chosen σ = (1−κ)1/2
and λ > (1 − κ)1/4n(d−1) . This gives condition (ii).
Finally, let z ∈ K. Recall that Lj = Dff j (x) Rj . Moreover, (Dhj )ϕj gj (z)
= Rj (we continue to identify Rj with its expression in the local chart ϕj ),
because
−1
g j (z) ∈ g −n+j (ϕ−1
n (Cn (yn , σρ))) ⊂ ϕj (Cj (yj , σρ)).
Therefore, writing zj = hj (ϕj (g j (z))) for simplicity,
Dggj (z) − Lj = D(fj )zj Rj − D(fj )0 Rj ≤ D(fj )zj − D(fj )0 Rj < γ.
The last inequality follows from our choice of r . This gives condition (iii) in
Definition 2.10. The proof of Lemma 3.3 is complete.
Remark 3.7. This last step explains why it is technically more convenient
to require Dggj (z) − Lj < γ, rather than Dggj (z) = Lj , when defining realizable sequence.
3.2. Proof of the directions interchange Proposition 3.1. First, we define
some auxiliary constants. Fix 0 < κ < 12 κ. Let ε1 > 0, depending on f , ε0
and κ , be given by Lemma 3.2. Let ε2 > 0, depending on f , √
ε0 and κ, be
given by Lemma 3.3. Take ε = min{ε1 , ε2 }. Fix α > 0 such that 2 sin α < ε.
Take
K ≥ (sin α)−2 and K ≥ max Dfx /m(Dfx ); x ∈ M .
(3.15)
Let β > 0 be such that
(3.16)
√
8 2 K sin β < ε sin6 α.
Finally, assume m ∈ N satisfies m ≥ 2π/β.
Let y ∈ M be a nonperiodic point and Ty M = E ⊕ F be a splitting as in
the hypothesis:
(3.17)
Dfym |F 1
≥ .
m
m(Dfy |E )
2
We write Ej = Dfyj (E) and Fj = Dfyj (F ) for j = 0, 1, . . . , m. The proof is
divided in three cases. Lemma 3.2 suffices for the first two; in the third step
we use the full strength of Lemma 3.3.
First case. Suppose there exists ∈ {0, 1, . . . , m} such that
(3.18)
(E , F ) < α.
LYAPUNOV EXPONENTS
1449
Fix as above. Take unit vectors ξ ∈ E and η ∈ F such that (ξ, η) < α.
⊥
Let Y = Rξ ⊕ Rη and X =
√ Y . Let R : Y → Y be a rotation such that
R(ξ) = η. Then R − I = 2 sin (ξ, η) < ε. Let R : Tf (y) M → Tf (y) M be
such that R preserves both X and Y , R|X = I and R|Y = R.
Consider first < m. By Lemma 3.2, the length 1 sequence {Dff (y) R}
is (κ , ε0 )-realizable at f (y). Using part 2 of Lemma 2.11 we conclude that
{L0 , . . . Lm−1 } = {Dfy , . . . , Dff −1 (y) , Dff (y) R, Dff +1 (y) , . . . , Dff m−1 (y) }
is a (κ, ε0 )-realizable sequence of length m at y. The case = m is similar.
By Lemma 3.2, the length 1 sequence {R Dff m−1 (y) } is (κ , ε0 )-realizable at
f m−1 (y). Then, by part 2 of Lemma 2.11,
{L0 , . . . Lm−1 } = {Dfy , . . . , Dff m−2 (y) , R Dff m−1 (y) }
is a (κ, ε0 )-realizable sequence of length m at y. In either case, Lm−1 . . . L0
sends the vector v = Df − (ξ) ∈ E0 to a vector w collinear to Df m− (η) ∈ Fm .
Second case. Assume there exist k, ∈ {0, . . . , m}, with k < , such that
Dff−k
k (y) |Fk (3.19)
m(Dff−k
k (y) |Ek )
> K.
Fix k and as above. Let ξ ∈ Ek , η ∈ Fk be unit vectors such that
Df −k (ξ) = m(Df −k |Ek )
and Df −k (η) = Df −k |Fk (Df −k is always meant at the point f k (y)). Define also unit vectors
ξ =
Df −k (ξ)
∈ E
Df −k (ξ)
and η =
Df −k (η)
∈ F .
Df −k (η)
: Rξ ⊕ Rη →
Let ξ1 = ξ + (sin α)η. Then θ = (ξ, ξ1 ) ≤ α. In particular, if R
Rξ ⊕ Rη is a rotation of angle ±θ, sending Rξ to Rξ1 then
√
− I = 2 sin θ < ε.
R
Let Y = Rξ ⊕ Rη and X = Y ⊥ . Let R : Tf k (y) M → Tf k (y) M be such that
By Lemma 3.2, the
R preserves both X and Y , with R|X = I and R|Y = R.
k
length 1 sequence {Dff k (y) R} is (κ , ε0 )-realizable at f (y). Let η1 = sξ + η ,
where
1 Df −k (ξ)
1 m(Df −k |Ek )
s=
=
.
−k
sin α Df (η)
sin α Df −k |Fk Then the vectors Df −k ξ1 and η1 are collinear. Besides, s < 1/(K sin α) <
sin α, because of (3.15) and (3.19). Hence, θ = (η1 , η) < α. Then, as
before, there exists R : Tf (y) M → Tf (y) M such that R (Rη1 ) = Rη and
{R Dff −1 (y) } is a (κ , ε0 )-realizable sequence of length 1 at f −1 (y).
1450
JAIRO BOCHI AND MARCELO VIANA
Notice that (3.15) and (3.19) imply − 1 > k. Then we may define a
sequence {L0 , . . . , Lm−1 } of linear maps as follows:

for j = k
 Dff k (y) R
Lj =
R Dff −1 (y) for j = − 1

Dff j (y)
for all other j.
By parts 1 and 2 of Lemma 2.11, this is a (κ, ε0 )-realizable sequence of length
m at y. By construction, Lm−1 . . . L0 sends v = Df −k (ξ) ∈ E0 to a vector w
collinear to Df m− (η ) ∈ Fm .
Third case.
we assume
We suppose that we are not in the previous cases, that is,
for every j ∈ {0, 1, . . . , m}, (Ej , Fj ) ≥ α,
(3.20)
and
(3.21)
for every i, j ∈ {0, . . . , m} with i < j,
Dffj−i
i (y) |Fi ≤ K.
m(Dffj−i
i (y) |Ei )
We now use the assumption (3.17), and the choice of m in (3.16). Take unit
vectors ξ ∈ E0 and η ∈ F0 such that Df m ξ = m(Df m |E0 ) and Df m η =
Df m |F0 (Df m is always computed at y). Let also η = Df m (η)/Df m (η)
∈ Fm .
Define G0 = E0 ∩ ξ ⊥ and Gj = Dfyj (G0 ) ⊂ Ej for 0 < j ≤ m. Dually,
define Hm = Fm ∩η ⊥ and Hj = Df j−m (Hm ) ⊂ Fj for 0 ≤ j < m. In addition,
⊥
consider unit vectors vj ∈ Ej ∩ G⊥
j and wj ∈ Fj ∩ Hj for 0 ≤ j ≤ m. These
vectors are uniquely defined up to a choice of sign, and v0 = ±ξ and wm = ±η .
See Figure 2.
Df j
G0
Gj
Gm
vm
ξ = v0
vj
Ej
E0
H0
w0
F0
Em
Fj
Hj
Fm
Hm
η = wm
wj
Df j−m
Figure 2: Setup for application of the nested rotations lemma
LYAPUNOV EXPONENTS
1451
For j = 0, . . . , m, define
X j = Gj ⊕ Hj
and Yj = Rvj ⊕ Rwj .
The spaces Xj are invariant: Dff j (y) (Xj ) = Xj+1 (the Yj are not). We
shall prove, using (3.20) and (3.21), that the maps Dfyj /X0 : Ty M/X0 →
Tf j (y) M/Xj do not distort angles too much:
Lemma 3.8. For every j = 0, 1, . . . , m,
Dfyj /X0 m(Dfyj /X0 )
≤
8K
.
sin6 α
Let us postpone the proof of this fact for a while, and proceed preparing the application Lemma 3.3. Let B0 ⊂ (Ty M )/X0 be a ball and Bj =
(Dfyj /X0 )(B0 ) for 0 < j ≤ m. Since mβ ≥ 2π, it is possible to choose numbers
θ0 , . . . , θm−1 such that |θj | ≤ β for all j and
(3.22)
m−1
θj = (v0 + X0 , w0 + X0 ).
j=0
Let Pj : (Ty M )/X0 → (Ty M )/X0 be the rotation of angle θj . Define linear
j : (Tf j (y) M )/Xj → (Tf j (y) M )/Xj by
maps R
j = Dfyj /X0 Pj Dfyj /X0 −1 .
R
j (Bj ) = Bj for all j. Moreover,
Since Pj preserves the ball B0 , we have R
Dfyj /X0 8K √
P
Rj − I ≤
−
I
≤
2 sin β < ε,
j
sin6 α
m(Dfyj /X0 )
√
by Lemma 3.8, the relation Pj − I ≤ 2 sin β, and our choice (3.16) of β.
j ) we obApplying Lemma 3.3 to these data (ε0 , κ, x = y, n = m, Xj , Bj , R
tain an (ε0 , κ)-realizable sequence {L0 , . . . , Lm−1 } at the point y, with Lj |Xj =
Dff j (y) |Xj and
j = (Df j+1 /Xj ) Pj (Df j /X0 )−1 .
Lj /Xj = (Dff j (y) /Xj ) R
y
y
Let L = Lm−1 . . . L0 . Then L/X0 = (Dfym /X0 )Pm−1 . . . P0 . In particular, by
(3.22),
L(v0 + X0 ) = (Dfym /X0 )(w0 + X0 ) = Dfym (w0 ) + Xm .
Recall that Xm = Gm ⊕ Hm by definition. Then we may write
L(v0 ) = Dfym (w0 ) + um + um
with um ∈ Gm and um ∈ Hm . Let u0 = (Dfym )−1 (um ) ∈ G0 ⊂ X0 ∩ E0 .
Since L equals Dfym on X0 , we have L(u0 ) = um . This means that the vector
1452
JAIRO BOCHI AND MARCELO VIANA
v = v0 − u0 ∈ E0 is sent by L to the vector Dfym (w0 ) + um ∈ Fm . This finishes
the third and last case of Proposition 3.1.
Now we are left to give the:
Proof of Lemma 3.8. Recall that Xj = Gj ⊕ Hj , Gj ⊂ Ej and Hj ⊂ Fj ,
vj ∈ Ej , wj ∈ Fj , and vj ⊥ Gj , wj ⊥ Hj . Hence, by (3.20),
(Xj , vj ) = (Hj , vj ) ≥ (Fj , Ej ) ≥ α
and
(Xj ⊕ Rvj , wj ) = (Rvj ⊕ Gj , wj ) ≥ (Ej , Fj ) ≥ α.
Using Lemma 2.6 with A = Xj , B = Rvj , C = Rwj , we deduce the following
lower bound for the angle between the spaces Xj and Yj = Rvj ⊕ Rwj :
sin (Xj , Yj ) ≥ sin (Xj , vj ) sin (Rvj ⊕ Xj , wj ) ≥ sin2 α.
Let πj : Yj → (Tf j (y) M )/Xj be the canonical map πj (w) = w + Xj . Then πj
is an isomorphism, πj = 1 and
(3.23)
πj−1 = 1/ sin (Yj , Xj ) ≤ 1/ sin2 α
(the quotient space has the norm that makes Xj⊥ w → w + Xj an isometry).
Now let pj : Tfj (y) M → Yj be the projection onto Yj associated to the
splitting Tfj (y) M = Xj ⊕ Yj . Let Dj : Yj → Yj+1 be given by Dj = pj+1 ◦
(Dff j (y) |Yj ). Define
D(j) : Y0 → Yj
by D(j) = Dj−1 ◦ · · · ◦ D0 = pj ◦ (Dfyj |Y0 ).
We claim that the following inequalities hold:
(3.24)
1
D(j) (w0 )
≤K
<
2K
D(j) (v0 )
for every j with 0 ≤ j ≤ m.
To prove this, consider the matrix of Dj relative to bases {vj , wj } and
{vj+1 , wj+1 }:
aj 0
.
Dj =
0 bj
Then D(j) (v0 ) = |aj−1 . . . a0 | and D(j) (w0 ) = |bj−1 . . . b0 | , since vj and wj
are unit vectors. Moreover, for 0 ≤ i < j ≤ m we have
−(j−i)
(vj )−1 ,
|aj−1 . . . ai | = pj ◦ Dffj−i
i (y) (vi ) = pi ◦ Dff j (y)
−(j−i)
(wj )−1 .
|bj−1 . . . bi | = pj ◦ Dffj−i
i (y) (wi ) = pi ◦ Dff j (y)
Recall that vs ∈ Es and ws ∈ Fs for all s. When restricted to Es (or Fs ), the
map ps is the orthogonal projection to the direction of vs (or ws ). In particular,
pi |Ei = pj |Fj = 1 and so
−(j−i)
|aj−1 . . . ai | ≥ Dff j (y) |Ej −1 = m(Dffj−i
i (y) |Ei )
LYAPUNOV EXPONENTS
1453
and
|bj−1 . . . bi | ≤ Dffj−i
i y |Fi .
Using (3.21), we obtain
(3.25)
|bj−1 . . . bi |
≤K
|aj−1 . . . ai |
for all 0 ≤ i ≤ j ≤ m.
Taking i = 0 gives the upper inequality in (3.24). For the same reasons, and
the definitions of v0 = ξ and wm = η = Dfym (η)/Dfym (η), we also have
|am−1 . . . a0 | ≤ Dfym (v0 ) = Dfym (ξ) = m(Df m |E0 ),
−1
= Dfym (η) = Df m |F0 .
|bm−1 . . . b0 | ≥ Dff−m
m (y) (wm )
Now (3.17) translates into
1
|bm−1 . . . b0 |
> .
|am−1 . . . a0 |
2
Combining this inequality and (3.25), with i, j replaced with j, m, we find
|bj−1 . . . b0 |
|bm−1 . . . b0 | |bm−1 . . . bj |
1/2
=
/
>
,
|aj−1 . . . a0 |
|am−1 . . . a0 | |am−1 . . . aj |
K
which is the remaining inequality in (3.24).
Now, combining Lemma 2.8 with (3.24) and (vs , ws ) ≥ α, we get
D(j) 8K
≤
.
m(D(j) )
sin2 α
Moreover, Dfyj /X0 = πj ◦ D(j) ◦ π0−1 . So, using the relation (3.23),
Dfyj /X0 m(Dfyj /X0 )
≤
πj π0 D(j) 8K
·
·
≤
.
m(πj ) m(D(j) ) m(π0 )
sin6 α
This finishes the proof of Lemma 3.8.
The proof of Proposition 3.1 is now complete.
4. Proofs of Theorems 1 and 2
Let us define some useful invariant sets. Given f ∈ Diff 1µ (M ), let O(f ) be
the set of the regular points, in the sense of the theorem of Oseledets. Given
p ∈ {1, . . . , d − 1} and m ∈ N, let Dp (f, m) be the set of points x such that
there is an m-dominated splitting of index p along the orbit of x. That is,
x ∈ Dp (f, m) if and only if there exists a splitting Tf n x M = En ⊕ Fn (n ∈ Z)
such that for all n ∈ Z, dim En = p, Dff n x (En ) = En+1 , Dff n x (Fn ) = Fn+1
and
Dffmn (x) |Fn 1
≤ .
m
m(Dff n (x) |En )
2
1454
JAIRO BOCHI AND MARCELO VIANA
By Section 2.2, Dp (f, m) is a closed set. Let
Γp (f, m) = M Dp (f, m),
Γp (f, m) = x ∈ Γp (f, m) ∩ O(f ); λp (f, x) > λp+1 (f, x) ,
Γ∗p (f, m) = x ∈ Γp (f, m); x is not periodic .
Define also
Γp (f, ∞) =
Γp (f, m)
and
Γp (f, ∞) =
m∈N
Γp (f, m).
m∈N
It is clear that all these sets are invariant under f .
Lemma 4.1. For every f and p, the set Γp (f, ∞) contains no periodic
points of f . In other words, m∈N Γp (f, m) Γ∗p (f, m) = ∅.
Proof. Suppose that x ∈ O(f ) is periodic, say, of period n, and λp (f, x) >
λp+1 (f, x). The eigenvalues of Dfxn are ν1 , . . . , νd , with |νi | = enλi (f,x) . Let E
(resp. F ) be the sum of the eigenspaces of Dfxn associated to the eigenvalues
ν1 , . . . , νp (resp. νp+1 , . . . , νd ). Then the splitting Tx M = E ⊕ F is Dfxn invariant. Spreading it along the orbit of x, we obtain a dominated splitting.
That is, x ∈ Dp (f, m) for some m ∈ N, and so x ∈ Γp (f, ∞).
4.1. Lowering the norm along an orbit segment. Recall that Λp (f, x) =
λ1 (x) + · · · + λp (x) for each p = 1, . . . , d.
Proposition 4.2. Let f ∈ Diff 1µ (M ), ε0 > 0, κ > 0, δ > 0 and p ∈
{1, . . . , d − 1}. Then, for every sufficiently large m ∈ N, there exists a measurable function N : Γ∗p (f, m) → N with the following properties: For almost
every x ∈ Γ∗p (f, m) and every n ≥ N (x) there exists an (ε0 , κ)-realizable se (x,n) , . . . , L
(x,n) } at x of length n such that
quence {L
0
n−1
1
(x,n) . . . L
(x,n) ) ≤ Λp−1 (f, x) + Λp+1 (f, x) + δ.
log ∧p (L
n−1
0
n
2
Proof. Fix f , ε0 , κ, δ and p. For clearness, we divide the proof into two
parts:
. Fix f , ε0 , κ, δ and p.
Part 1. Definition of N (·) and the sequence L
j
Assume m ∈ N is sufficiently large so that the conclusion of Proposition 3.1
holds for f , ε0 and 12 κ (in the place of κ). To simplify the notation, let Γ =
Γ∗p (f, m). We may suppose that µ(Γ) > 0; otherwise there is nothing to prove.
Consider the splitting TΓ M = E ⊕ F , where E is the sum of the Oseledets
subspaces corresponding to the first p Lyapunov exponents λ1 ≥ · · · ≥ λp and
F is the sum of the subspaces corresponding to the other exponents λp+1 ≥
· · · ≥ λd . This makes sense since λp > λp+1 on Γ. Let A ⊂ Γ be the set of
(x,n)
LYAPUNOV EXPONENTS
1455
points y such that the nondomination condition (3.17) holds. By definition of
Γ = Γ∗p (f, m),
(4.1)
f n (A).
Γ=
n∈Z
d
Let λ∧p
i (x), 1 ≤ i ≤ p , denote the Lyapunov exponents of the cocycle
p
∧ (Df ) over f , in nonincreasing order. Let Vx denote the Oseledets subspace
associated to the upper exponent λ∧p
1 (x) and let Hx be the sum of all other
Oseledets subspaces. This gives us a splitting ∧p (T M ) = V ⊕ H. By Proposition 2.1,
λ∧p
1 (x) = λ1 (x) + · · · + λp−1 (x) + λp (x),
λ∧p
2 (x) = λ1 (x) + · · · + λp−1 (x) + λp+1 (x).
∧p
If x ∈ Γ then λp (x) > λp+1 (x) and so λ∧p
1 (x) > λ2 (x). That is, the subspace
Vx is one-dimensional.
For almost every x ∈ Γ, Oseledets’ theorem gives Q(x) ∈ N such that for
all n ≥ Q(x), we have:
•
1
n
log ∧
p
(Dfxn )v
v
< λ∧p
1 (x) + δ for every v ∈ Vx {0};
•
1
n
log ∧
p
(Dfxn )w
w
< λ∧p
2 (x) + δ for every w ∈ Hx {0};
•
1
n
log sin (Vf n x , Hf n x ) > −δ.
For q ∈ N, let Bq = {x ∈ Γ; Q(x) ≤ q}. Then Bq ↑ Γ; that is, the Bq form a
nondecreasing sequence and their union is a full measure subset of Γ. Define
C0 = ∅ and
(4.2)
f n (A ∩ f −m (Bq )).
Cq =
n∈Z
Since f −m (Bq ) ↑ Γ and the definition (4.1), we have Cq ↑ Γ. To prove Proposition 4.2 we must define the function N on Γ. We are going to define it on
each of the sets Cq Cq−1 separately. From now on, let q ∈ N be fixed.
We need the following recurrence result, proved in [4, Lemma 3.12].
Lemma 4.3. Let f ∈ Diff 1µ (M ). Let A ⊂ M be a measurable set with
µ(A) > 0, and let Γ = ∪n∈Z f n (A). Fix any γ > 0. Then there exists a
measurable function N0 : Γ → N such that for almost every x ∈ Γ, and for all
n ≥ N0 (x) and t ∈ (0, 1), there exists ∈ {0, 1, . . . , n} such that t − γ ≤ /n ≤
t + γ and f (x) ∈ A.
Let c be a strict upper bound for log ∧p (Df ) and γ = min{c−1 δ, 1/10}.
(q)
Using (4.2) and Lemma 4.3, we find a measurable function N0 : Cq → N such
1456
JAIRO BOCHI AND MARCELO VIANA
(q)
that for almost every x ∈ Cq , every n ≥ N0 (x) and every t ∈ (0, 1) there is
∈ {0, 1, . . . , n} with |
/n − t| < γ and f x ∈ A ∩ f −m (Bq ). We define N (x)
for x ∈ Cq Cq−1 as the least integer such that
N (x) ≥ max{N0 (x), 10Q(x), mγ −1 , δ −1 log[4/ sin (Vx , Hx )]}.
(q)
Now fix a point x ∈ Cq Cq−1 and n ≥ N (x). We will now construct the
(x,n) }. Since n ≥ N (q) (x), there exists ∈ N such that
sequence {L
0
j
− 1 < γ and y = f (x) ∈ A ∩ f −m (Bq ).
n 2
Since y ∈ A, where the nondomination condition (3.17) holds, Proposition 3.1
gives a sequence {L0 , . . . , Lm−1 } which is (ε0 , 12 κ)-realizable , such that there
are nonzero vectors v0 ∈ Ey , w0 ∈ Ff m (y) for which
Lm−1 . . . L0 (v0 ) = w0 .
We form the sequence {L
0
(x,n)
{Dff i (x) ; 0 ≤ i < },
,...,L
n−1 } of length n by concatenating
(x,n)
{L0 , . . . , Lm−1 },
{Dff i (x) ; + m ≤ i < m}.
According to parts 1 and 2 of Lemma 2.11, the concatenation is an (ε0 , κ)realizable sequence at x.
(x,n) . . . L
(x,n) ). Write ∧p (L
(x,n) . . . L
(x,n) ) =
Part 2. Estimation of ∧p (L
n−1
0
n−1
0
p
D1 LD0 , with D0 = ∧p (Dfx ), D1 = ∧p (Dffn−−m
+m (x) ), and L = ∧ (Lm−1 . . . L0 ).
The key observation is:
Lemma 4.4. The map L : ∧p (Ty M ) → ∧p (Tf m (y) M ) satisfies L(Vy ) ⊂
Hf m (y) .
Proof. Proposition 2.1 describes the spaces V and H. Let z ∈ Γ and
consider a basis {e1 (z), . . . ed (z)} of Tz M such that
ei (x) ∈ Exj
for dim Ex1 + · · · + dim Exj−1 < i ≤ dim Ex1 + · · · + dim Exj .
Then Vz is the space generated by e1 (z) ∧ · · · ∧ ep (z) and Hz is generated by
the vectors ei1 (z) ∧ · · · ∧ eip (z) with 1 ≤ i1 < · · · < ip ≤ d, ip > p. Also notice
that {e1 (z), . . . , ep (z)} and {ep+1 (z), . . . , ed (z)} are bases for the spaces Ez and
Fz , respectively. Consider the vectors v0 ∈ Ey and w0 = L(v0 ) ∈ Ff m y , where
L = Lm−1 . . . L0 . There is ν ∈ {1, . . . , p} such that
{v0 , e1 (y), . . . , eν−1 (y), eν+1 (y), . . . , ep (y)}
is a basis for Ey . Therefore Vy is generated by the vector
v0 ∧ e1 (y) ∧ · · · ∧ eν−1 (y) ∧ eν+1 (y) ∧ · · · ∧ ep (y),
LYAPUNOV EXPONENTS
1457
which is mapped by L = ∧p (L) to
(4.3)
w0 ∧ Le1 (y) ∧ · · · ∧ Leν−1 (y) ∧ Leν+1 (y) ∧ · · · ∧ Lep (y),
Write w0 as a linear combination of vectors ep+1 (f m (y)), . . . , ed (f m (y)) and
write each Lei (y) as a linear combination of vectors e1 (f m (y)), . . . , ed (f m (y)).
Substituting in (4.3), we get a linear combination of ei1 (f m (y))∧· · ·∧eip (f m (y))
where e1 (f m (y)) ∧ · · · ∧ ep (f m (y)) does not appear. This proves that the vector
in (4.3) belongs to Hf m (y) .
To carry on the estimates, we introduce a more convenient norm: For x0 ,
x1 ∈ Γ we represent a linear map T : ∧p (Tx0 M ) → ∧p (Tx1 M ) by its matrix
++
T
T +−
T =
T −+ T −−
with respect to the splittings Tx0 M = Vx0 ⊕ Hx0 and Tx1 M = Vx1 ⊕ Hx1 . Then
we define
T max = max T ++ , T +− , T −+ , T −− .
The following elementary lemma relates this norm to the original one T (that
comes from the metric in ∧p (TΓ M ) ).
Lemma 4.5. Let θx0 = (Vx0 , Hx0 ) and θx1 = (Vx1 , Hx1 ). Then:
(1) T ≤ 4 (sin θx0 )−1 T max ;
(2) T max ≤ (sin θx1 )−1 T .
Proof. Let v = v+ + v− ∈ Vx0 ⊕ Hx0 . We have v∗ ≤ v/ sin θx0 for
∗ = + and ∗ = −. Thus,
T v ≤ T ++ v+ + T ++ v− + T −− v+ + T −− v− ≤ 4T max v/ sin θx0 .
This proves part 1. The proof of part 2 is similar. Let v+ ∈ Vx0 . Its image
splits as T v+ = T ++ v+ + T −+ v+ ∈ Vx1 ⊕ Hx1 . Hence,
T ∗+ v+ ≤ T v+ (sin θx1 )−1 ≤ T v+ (sin θx1 )−1
for ∗ = + and ∗ = −. Together with a corresponding estimate for T ∗− v− ,
v− ∈ Hx0 , this gives part 2.
For the linear maps we were considering, the matrices have the form:
++
0
0
L+−
Di
, i = 0, 1, and L =
:
Di =
0
Di−−
L−+ L−−
Di+− = 0 = Di−+ because V and H are ∧p (Df )-invariant, and L++ = 0
because of Lemma 4.4. Then
0
D1++ L+− D0−−
p (4.4)
.
∧ (Ln−1 . . . L0 ) =
D1−− L−+ D0++ D1−− L−− D0−−
1458
JAIRO BOCHI AND MARCELO VIANA
Lemma 4.6. For i = 0, 1, x ∈ Cq Cq−1 and n ≥ N (x),
log Di++ < 12 n(λ∧p
1 (x) + 5δ)
Proof. Since > ( 12 − γ)n >
and
log Di−− < 12 n(λ∧p
2 (x) + 5δ).
1
10 N (x) ≥ Q(x), we have
log D0++ = log ∧ (Dfx )|Vx < (λ∧p
1 (x) + δ),
−−
p
log D0 = log ∧ (Dfx )|Hx < (λ∧p
2 (x) + δ).
∧p
Let λ be either λ∧p
1 (x) or λ2 (x). Using γλ < γc ≤ δ and γ < 1, we find
(λ + δ) < n( 12 + γ)(λ + δ) < n( 12 λ + 12 δ + δ + δ) = 12 n(λ + 5δ).
1
n
This proves the case i = 0. We have n − − m > n( 12 − γ) − nγ > 10
−m
+m
Q(x) ≥ q. Also f (x) ∈ f (Bq ), and so Q(f
(x)) ≤ q. Therefore
++
n−−m
log D1 = log ∧p (Dff +m x )|Vf +m x < (n − − m)(λ∧p
1 (x) + δ),
−−
n−−m
p
log D1 = log ∧ (Dff +m x )|Hf +m x < (n − − m)(λ∧p
2 (x) + δ).
1
10 n
p
≥
≥
As before, (n − − m)(λ + δ) < n( 12 + γ)(λ + δ) 12 n(λ + 5δ). This proves the
case i = 1.
Lemma 4.7. log Lmax < 2nδ.
Proof. Since the sequence {L0 , . . . , Lm−1 } is realizable, each Lj is close to
the value of Df at some point. Therefore we may assume that log ∧p (Lj ) < c.
1
n≥
In particular, log L < mc ≤ ncγ ≤ nδ. We have + m ≥ n( 12 − γ) ≥ 10
Q(x). So log[1/ sin (Vf +m x , Hf +m x )] < δ and, by part 2 of Lemma 4.5,
log Lmax < 2nδ.
Using Lemmas 4.6 and 4.7, we bound each of the entries in (4.4):
∧p
log D1++ L+− D0−− < 12 n(λ∧p
1 (x) + λ2 (x) + 14δ),
∧p
log D1−− L−+ D0++ < 12 n(λ∧p
1 (x) + λ2 (x) + 14δ),
log D1−− L−− D0−− < 12 n(2λ∧p
2 (x) + 14δ).
The third expression is smaller than either of the first two, and so we get
∧p
∧p
n−1 . . . L
0 )max < n λ1 (x) + λ2 (x) + 7δ .
log ∧p (L
2
Therefore, by part 1 of Lemma 4.5 and log[4/ sin (Vx , Hx )] < nδ,
∧p
∧p
n−1 . . . L
0 ) < n λ1 (x) + λ2 (x) + 8δ .
log ∧p (L
2
∧p
∧p
We also have λ1 (x) + λ2 (x) = Λp−1 (f, x) + Λp+1 (f, x). This proves Proposition 4.2 (replace δ with δ/8 in the proof).
4.2. Globalization. The following proposition renders global the construction of Proposition 4.2:
1459
LYAPUNOV EXPONENTS
Proposition 4.8. Let f ∈ Diff 1µ (M ), ε0 > 0, p ∈ {1, . . . , d − 1} and
δ > 0. Then there exist m ∈ N and a diffeomorphism g ∈ U(f, ε0 ) that equals
f outside the open set Γp (f, m) such that
Λp−1 (f, x) + Λp+1 (f, x)
Λp (g, x) dµ(x) < δ +
dµ(x).
2
Γp (f,m)
Γp (f,m)
We need some preparatory terminology:
Definition 4.9. Let f ∈ Diff 1µ (M ). An f -tower (or simply tower ) is a
pair of measurable sets (T, Tb ) such that there is a positive integer n, called
the height of the tower, such that the sets Tb , f (Tb ), . . . , f n−1 (Tb ) are pairwise
disjoint and their union is T . Now, Tb is called the base of the tower.
An f -castle (or simply castle) is a pair of measurable sets (Q, Qb ) such
that there exists a finite or countable family of pairwise disjoint towers (Ti , Tib )
such that Q = Ti and Qb = Tib and Qb is called the base of the castle.
A castle (Q, Qb ) is a sub-castle of a castle (Q , Qb ) if Qb ⊂ Qb and for
every point x ∈ Qb , the towers of (Q, Qb ) and (Q , Qb ) that contain x have
equal heights. In particular, Q ⊂ Q .
We shall frequently omit reference to the base of a castle Q in our notation.
Definition 4.10. Given f ∈ Diff 1µ (M ) and a positive measure set A ⊂ M ,
consider the return time τ : A → N defined by τ (x) = inf{n ≥ 1; f n (x) ∈ A}.
If we denote An = τ −1 (n) then Tn = An ∪ f (An ) ∪ · · · ∪ f n−1 (An ) is a tower.
Consider the castle Q, with base A, given by the union of the towers Tn . This
is called the Kakutani castle with base A.
Note that Q = n∈Z f n (A) mod 0; in particular the set Q is invariant.
Proof of Proposition 4.8. Let f , ε0 , p and δ be given. For simplicity, we
write
Λp−1 (f, x) + Λp+1 (f, x)
φ(x) =
.
2
i ⊃ Qi . Let κ = δ 2 . Take
Step 1. Construction of families of castles Q
m ∈ N large enough so that the conclusion of Proposition 4.2 holds: There exists a measurable function N : Γ∗p (f, m) → N such that for a.e. x ∈ Γ∗p (f, m) and
(x,n) , . . . , L
(x,n) }
every n ≥ N (x) there exists an (ε0 , κ)-realizable sequence {L
0
at x of length n such that
1
(x,n) . . . L
(x,n) ) ≤ φ(x) + δ.
(4.5)
log ∧p (L
n−1
0
n
We shall also (see Lemma 4.1) assume that m is large enough so that
µ Γp (f, m) Γ∗p (f, m) < δ.
(4.6)
n−1
1460
JAIRO BOCHI AND MARCELO VIANA
Let C > supg∈U (f,ε0 ) supy∈M log Dgy p and = C/δ. For i = 1, 2, . . . , , let
Z i = {x ∈ Γ∗p (f, m); (i − 1)δ ≤ φ(x) < iδ}.
Each Z i is an f -invariant set. Since φ < C, we have Γ∗p (f, m) = i=1 Z i .
Define the sets Zni = {x ∈ Z i ; N (x) ≤ n} for n ∈ N and 1 ≤ i ≤ . Obviously,
Zni ↑ Z i when n → ∞. Fix H ∈ N such that, for all i = 1, 2, . . . , ,
(4.7)
i
µ(Z i ZH
) < δ 2 µ(Z i ).
Using the fact that Λp (f ) equals φ in the f -invariant set Γp (f, m) Γp (f, m),
and Proposition 2.2, we may also assume that H is large enough so that
1
p
n
(4.8)
φ
log ∧ (Df ) < δ +
Γp (f,m)Γp (f,m) n
Γp (f,m)Γp (f,m)
for all n ≥ H.
A measure-preserving transformation is aperiodic if the set of periodic
points has zero measure. The following result was proved in [4, Lemma 4.1]:
Lemma 4.11. For every aperiodic invertible measure-preserving transformation f on a probability space X, every subset U ⊂ X of positive measure,
and every n ∈ N, there exists a positive measure set V ⊂ U such that the
sets V , f (V ), . . . , f n (V ) are two-by-two disjoint. Besides, V can be chosen to
be maximal in the measure-theoretical sense: no set that includes V and has
larger measure than V has the stated properties.
By definition of the set Γ∗p (f, m), the map f : Γ∗p (f, m) → Γ∗p (f, m)
i such that
is aperiodic. So, by Lemma 4.11, for each i there is B i ⊂ ZH
i
i
H−1
i
i
B , f (B ), . . . , f
(B ) are two-by-two disjoint and such that B is maximal
for these properties (in the measure-theoretical sense). Consider the following
f -invariant set:
i =
Q
f n (B i ).
n∈Z
i is the Kakutani castle with base B i . It is contained in Z i and, by the
Q
i up to a zero-measure subset. Thus, by (4.7),
maximality of B i , it contains ZH
(4.9)
i ) < δ 2 µ(Z i ).
µ(Z i Q
i be the sub-castle consisting of all the towers of Q
i with heights
Let Qi ⊂ Q
at most 3H floors. The following is a key property of the construction:
Lemma 4.12. For each i
i Qi ) ≤ 3µ(Z i Z i ).
µ(Q
H
=
1, 2, . . . , , there exists the relation
i into towers as
Proof. We follow [4, Lemma 4.2] and split the castle Q
k−1 j i
∞
∞
i
i
i
i
i =
i
Q
k=H Tk where B =
k=H Bk is the base Qb and Tk =
j=0 f (Bk )
1461
LYAPUNOV EXPONENTS
is the tower with base Bki and of height k floors. Take k ≥ 2H and H ≤
j ≤ k − H. The sets f j (Bki ), . . . f j+H−1 (Bki ) are disjoint and do not intersect
B i · · · f H−1 (B i ). Since B i is maximal, we conclude that
k ≥ 2H and H ≤ j ≤ k − H
⇒
i
µ(f j (Bki ) ∩ ZH
)=0
i ), contradicting the
(otherwise we could replace B i with B i (f j (Bki ) ∩ ZH
i
maximality of B ). Thus
k ≥ 2H
⇒
µ(Tki
i
ZH
)
≥
k−H
µ(f j (Bki )) =
j=H
In particular,
k − 2H + 1
µ(Tki ).
k
1
i
) > µ(Tki )
k ≥ 3H + 1 ⇒ µ(Tki ZH
3
and so
∞
i Qi ) =
µ(Q
µ(Tki ) ≤
k=3H+1
∞
i
3µ(Tk ZH
)
k=3H+1
!
#
∞
"
= 3µ
Tki
i
ZH
i
≤ 3µ(Z i ZH
),
k=3H+1
as claimed.
Step 2. Construction of the diffeomorphism g.
Lemma 4.13. For almost every x ∈ Γ∗p (f, m) and every n ≥ N (x), there
exists r > 0 such that for every ball U = Br (x) with 0 < r < r there exist
h ∈ U(f, ε0 ) and a measurable set K ⊂ Br (x) such that
j
(i) h equals f outside n−1
j=0 f (Br (x));
(ii) µ(K) > (1 − κ)µ(Br (x));
(iii) if y ∈ K then
1
n
log ∧p (Dhny ) < φ(x) + 2δ.
Proof. Fix x and n ≥ N (x). Recall that the point x is not periodic. Let
(x,n) } given by Proposition 4.2 is
γ > 0 be very small. Since the sequence {L
j
(κ, ε0 )-realizable, there exists r > 0 such that for every ball U = Br (x) with
0 < r < r there exists h ∈ U(f, ε0 ) satisfying condition (i) above and there
exists K ⊂ Br (x) satisfying condition (ii) and
y ∈ K and 0 ≤ j ≤ n − 1
⇒
(x,n) < γ.
Dhhj y − L
j
Taking γ small enough, this inequality and (4.5) imply
1
1
(x,n) . . . L
(x,n) ) + δ ≤ φ(x) + 2δ,
y∈K ⇒
log ∧p (Dhny ) < log ∧p (L
n−1
0
n
n
as claimed in the lemma.
1462
JAIRO BOCHI AND MARCELO VIANA
Lemma 4.14. Fix γ > 0. There exists g ∈ U(f, ε0 ) and for each i =
1, 2, . . . , there exist a g-castle U i and a g-sub-castle K i such that:
(i) the U i are open, pairwise disjoint, and contained in Γp (f, m);
(ii) µ(U i Qi ) < 2γµ(Z i ) and µ(Qi U i ) < 2γµ(Z i );
(iii) µ(U i K i ) < 2κµ(Z i );
(iv) g(U i ) = f (U i ) and g equals f outside
i=1 U
i;
(v) if y is in the base of K i and n(y) is the height of the tower of K i that
contains x then
1
log ∧p (Dgyn(y) ) < iδ + 2δ.
n(y)
Proof. By the regularity of the measure µ, one can find a compact subcastle J i ⊂ Qi such that
(4.10)
i ).
µ(Qi J i ) < γµ(Q
Since the J i are compact and disjoint we can find open pairwise disjoint castles
V i such that each V i contains J i as a sub-castle, is contained in the open and
invariant set Γp (f, m), and
(4.11)
i ).
µ(V i J i ) < γµ(Q
For each x ∈ Jbi , let n(x) be the height of the tower that contains x. Jbi
i , so that N (x) ≤ H ≤ n(x). Let r(x) > 0 be the radius
is contained in ZH
given by Lemma 4.13, with n = n(x). This is defined for almost every x ∈ Jbi .
Reducing r(x) if needed, we suppose that the ball B r(x) (x) is contained in the
base of a tower in V i (with the same height).
Using Vitali’s covering lemma1 , we can find a finite collection of disjoint
balls Uki = Brk,i (xk,i ), with xk,i ∈ Jbi and 0 < rk,i < r(xk,i ), such that
" µ Jbi Uki < γµ(Jbi ).
(4.12)
k
Let nk,i = n(xk,i ). Notice that n(x) = nk,i for all xi ∈ Uki .
Now we apply Lemma 4.13 to each ball Uki . We get, for each k, a measurable set Kki ⊂ Uki and a diffeomorphism hk,i ∈ U(f, ε0 ) such that (in 3 we use
the fact that xk,i ∈ Z i )
nik −1 j i
(1) hk,i equals f outside the set j=0
f (Uk );
i
i
(2) µ(Kk ) > (1 − κ) µ(Uk );
n
(3) if y ∈ Kki then n1k,i log ∧p (Dhk,ik,i )y < φ(xk,i ) + 2δ < iδ + 2δ.
1
First, cover the basis Jbi of the castle by chart domains.
1463
LYAPUNOV EXPONENTS
nik −1 j k
Let g be equal to hk,i in the set j=0
f (Ui ), for each i and k, and be
equal to f outside. Since those sets are disjoint, g ∈ Diff 1µ (M ) is a well-defined
diffeomorphism. Each hk,i belongs to U(f, ε0 ) and so g also does.
Since each Uki is contained in the base of a tower in the castle V i , V i is also
a castle for g. Let U i be the g-sub-castle of V i with base k Uki . Analogously,
let K i be the g-sub-castle of U i with base k Kki .
It remains to prove claims (ii) and (iii) in the lemma. Making use of the
castle structures, relation (4.12) and item 2 above imply, respectively,
(4.13)
µ(J i U i ) < γµ(J i )
and µ(U i K i ) < κµ(U i ).
i ⊂ Z i ,
By (4.11) and Q
(4.14)
i ) ≤ γµ(Z i ).
µ(U i Qi ) < µ(V i J i ) < γµ(Q
This implies the first part of item (ii). Combining the first part of (4.13)
with (4.10),
i ) ≤ 2γµ(Z i ).
µ(Qi U i ) < µ(Qi J i ) + µ(J i U i ) < 2γµ(Q
This proves the second part of item (ii). Finally, the second inequality in (4.13)
and
i ) < 2µ(Z i ).
µ(U i ) < µ(Qi ) + µ(U i Qi ) < (1 + γ)µ(Q
imply item (iii). The lemma is proved.
Step 3. Conclusion of the proof of Proposition 4.8. Let U = i=1 U i and
= Q
i . Set N = Hδ −1 . (Of course, we can assume
Q = i=1 Qi and Q
i=1
−1
that δ ∈ N.) Let
G=
"
G
i
where
i=1
G =Z ∩
i
i
N
−1
g −j (K i )
j=0
for each i = 1, 2, . . . , . The next lemma means that on G we managed to
reduce some time N exponent:
Lemma 4.15. If x ∈ Gi then
1
log ∧p (DgxN ) < iδ + (6C + 2)δ.
N
Proof. For y ∈ Kbi , let n(y) be the height of the g-tower containing y.
Take x ∈ G; say x ∈ Gi . Since the heights of towers of K i are less than 3H,
we can write
N = k1 + n1 + n2 + · · · + nj + k2
so that 0 ≤ k1 , k2 < 3H, 1 ≤ n1 , . . . , nj < 3H, and the points
x1 = g k1 (x), x2 = g k1 +n1 (x), . . . , xj+1 = g k1 +n1 +···+nj (x)
1464
JAIRO BOCHI AND MARCELO VIANA
are exactly the points (of the orbit segment x, g(x), . . . , g N −1 (x)) which belong
to Kbi . Write
2
∧p (DgxN ) ≤ ∧p (Dgxk1 ) ∧p (Dgxn11 ) . . . ∧p (Dgxnjj ) ∧p (Dgxkj+1
).
Using item (v) of Lemma 4.14, and our choice of N = Hδ −1 , we get
log ∧p (DgxN ) < k1 C + (n1 + · · · + nj )(iδ + 2δ) + k2 C
< 6HC + N (iδ + 2δ) < N (6Cδ + iδ + 2δ),
as claimed.
We also use the fact that G covers most of U ∪ Γ∗p (f, m), as asserted by
the next lemma whose proof we postpone for a while.
Lemma 4.16. Let γ = δ 2 /(
H) as in Lemma 4.14. Then
µ U ∪ Γ∗p (f, m) G < 12δ.
Continuing with the proof of Proposition 4.8, write ψ(x) = N1 log ∧p (DgxN ).
Since g leaves invariant the set Γp (f, m), Proposition 2.2 gives
Λp (g) ≤
ψ.
Γp (f,m)
Γp (f,m)
We split the integral on the right-hand side as
ψ=
ψ+
Γp (f,m)(U ∪Γp (f,m))
Γp (f,m)
(U ∪Γp (f,m))G
ψ+
ψ
G
= (I) + (II) + (III).
Outside U , g equals f and so ψ equals N1 log ∧p (Df N ). Thus
1
(I) ≤
φ,
log ∧p (Df N ) < δ +
Γp (f,m)Γp (f,m) N
Γp (f,m)Γp (f,m)
by (4.8). By Lemma 4.16 and (4.6), µ (U ∪Γp (f, m))G < 13δ. Since ψ < C,
we have (II) ≤ 13Cδ. By Lemma 4.15,
(III) =
i=1
Gi
ψ≤
(iδ + (6C + 2)δ)µ(Gi ) < (6C + 3)δ +
i=1
(i − 1)δµ(Gi ).
i=1
Since φ ≥ (i − 1)δ inside Z i ⊃ Gi , we have
(III) < (6C + 3)δ +
φ.
Γ∗p (f,m)
Summing the three terms, we get the conclusion of Proposition 4.8 (replace δ
1465
LYAPUNOV EXPONENTS
with δ/(18C + 4) throughout the arguments):
Λp (g) < (18C + 4)δ +
Γp (f,m)
φ.
Γp (f,m)
This completes the proof of the proposition, modulo proving Lemma 4.16.
Step 4. Proof of Lemma 4.16. The following observations will be useful
in the proof: If X ⊂ M is a measurable set and N ∈ N, then
−1
N
µ
g −j (X) ≤ µ(X) + (N − 1)µ g −1 (X) X .
(4.15)
j=0
Moreover, µ g −1 (X) X = µ X g −1 (X) .
Proof of Lemma 4.16. We shall prove first that
i Gi ) < 10δµ(Z i ).
µ(Q
i Gi ⊂ Q
i ∩ N −1 g −j (M K i ). Substituting
i ⊂ Z i , we have Q
Since Q
j=0
(4.16)
i Qi ) ∪ (M Q
i ),
M K i ⊂ (U i K i ) ∪ (Qi U i ) ∪ (Q
we obtain
i Gi ⊂
Q
N
−1
g −j (U i K i ) ∪
j=0
j=0
$
∪
N
−1
i ∩
Q
N
−1
g −j (Qi U i ) ∪
%
N
−1
i Qi )
g −j (Q
j=0
i ) = (I) ∪ (II) ∪ (III) ∪ (IV).
g −j (M Q
j=1
Let us bound the measure of each of these sets. The second one is easy: by
Lemma 4.14(ii) and our choices γ = δ 2 /
H and N = H/δ,
µ(II) ≤ N µ(Qi U i ) < 2N γµ(Z i ) < δµ(Z i ).
The other terms are more delicate.
The set X1 = U i K i is a g-castle whose towers have heights at least H.
Hence its base, which contains X1 g(X1 ), measures at most H1 µ(X1 ). By (4.15),
we get
N
µ(X1 ) < 2δ −1 µ(X1 ).
µ(I) < 1 +
H
By Lemma 4.14(iii), we have µ(X1 ) < 2κµ(Z i ) = 2δ 2 µ(Z i ). So, µ(I) <
4δµ(Z i ).
i Qi . By Lemma 4.12 and (4.7), we have µ(X3 ) < δ 2 µ(Zi ).
Let X3 = Q
Since f and g differ only in U , we have
g(X3 ) X3 ⊂ [f (X3 ) X3 ] ∪ g(X3 ∩ U ) = (V) ∪ (VI).
1466
JAIRO BOCHI AND MARCELO VIANA
Since X3 is an f -castle whose towers have heights of at least 3H,
µ(V) = µ(X3 f (X3 )) ≤
1
µ(X3 ).
3H
Since X3 ∩ U ⊂ k (U k Qk ), Lemma 4.14(ii) gives µ(VI) ≤ 2
γµ(Z i ). Combining the estimates of µ(V), µ(VI), µ(X3 ) with (4.15) and the definitions of
N and γ, we have
1
µ(III) < µ(X3 ) + N
µ(X3 ) + 2
γµ(Z i )
3H
1
µ(X3 ) + 2δµ(Z i ) < 3δµ(Z i ).
< 1+
3δ
We also have
i (IV) = Q
N
−1
g
−j
i ) ⊂
(Q
j=1
N
−1
i ) g −j (Q
i ) .
g j−1 (Q
j=1
i )). Notice that Q
i g −1 (Q
i ) ⊂
i g −1 (Q
In particular, µ(IV) ≤ (N − 1)µ(Q
k
i
k U (since Q is f -invariant). Therefore
i ) ⊂ [Q
i ∩ k=i U k ] ∪ [U i g −1 (Q
i )] = (VII) ∪ (VIII).
i g −1 (Q
Q
Combining
(VII) ⊂
"
k ) ⊂
(U k Q
k=i
"
(U k Qk )
k=i
with Lemma 4.14(ii) we obtain µ(VII) ≤ 2(
− 1)γµ(Z i ). Using the fact that
i = f (Q
i ), we also get
g(U i ) = f (U i ) and Q
i ) = µ(U i Q
i ) ≤ µ(U i Qi ) < 2γµ(Z i ).
µ(VIII) = µ(g(U i ) Q
i )) < 2
γµ(Z i ) and µ(IV) ≤ 2N γµ(Q
i ) < 2δµ(Zi ).
i g −1 (Q
Altogether, µ(Q
Summing the four parts, we obtain (4.16). Now
+ µ(U Q)
+ µ(Q
G)
µ U ∪ Γ∗p (f, m) G ≤ µ(Γ∗p (f, m) Q)
= µ(IX) + µ(X) + µ(XI).
Using (4.9), Lemma 4.14, and (4.16), respectively, we get
i ) < δ 2 < δ,
µ(Z i Q
µ(IX) ≤
i
µ(X) ≤ µ(U Q) ≤
µ(XI) ≤
i ) < 2γ < δ,
µ(U i Q
i
i G) < 10δ.
µ(Q
i
Summing the three parts, we conclude the proof of Lemma 4.16.
1467
LYAPUNOV EXPONENTS
4.3. End of the proofs of Theorems 1 and 2. We give an explicit lower
bound for the discontinuity “jump” of the semi-continuous function LEp (·).
Denote, for each p = 1, . . . , d,
λp (f, x) − λp+1 (f, x)
Jp (f ) =
dµ(x).
2
Γp (f,∞)
Proposition 4.17. Given f ∈ Diff 1µ (M ) and p ∈ {1, . . . , d − 1}, and
given any ε0 > 0 and δ > 0, there exists a diffeomorphism g ∈ U(f, ε0 ) such
that
Λp (g, x) dµ(x) <
Λp (f, x) dµ(x) − Jp (f ) + δ.
M
M
Proof. Let f , p, ε0 and δ be as in the statement. Using Proposition 4.8,
we find m ∈ N and g ∈ U(f, ε0 ) such that g = f outside Γp (f, m) and
Λp−1 (f ) + Λp+1 (f )
Λp (g) < δ +
.
2
Γp (f,m)
Γp (f,m)
Then
Λp (g) =
Λp (g) +
Γp (f,m)
M
<δ+
Γp (f,m)
M Γp (f,m)
Λp (g)
Λp−1 (f ) + Λp+1 (f )
+
2
M Γp (f,m)
Λp (f ).
Since Γp (f, ∞) ⊂ Γp (f, m), and the integrand is nonnegative,
Λp−1 (f ) + Λp+1 (f )
Λp (f ) −
2
Γp (f,m)
Λp−1 (f ) + Λp+1 (f )
= Jp (f ).
Λp (f ) −
≥
2
Γp (f,∞)
Therefore, the previous inequality implies
Λp (g) < δ − Jp (f ) +
M
Λp (f ),
M
as we wanted to prove.
Theorem 2 follows easily from Proposition 4.17:
Proof of Theorem 2. Let f ∈ Diff 1µ (M ) be a point of continuity of LEp (·)
for all p = 1, . . . , d−1. Then Jp (f ) = 0 for every p. This means that λp (f, x) =
λp+1 (f, x) for almost every x ∈ Γp (f, ∞). Let x ∈ M be an Oseledets regular
point. If all Lyapunov exponents of f at x vanish, there is nothing to prove.
Otherwise, for any p such that λp (f, x) > λp+1 (f, x), the point x ∈
/ Γp (f, ∞)
(except for a zero-measure set of x) . This means that x ∈ Dp (f, m) for
1468
JAIRO BOCHI AND MARCELO VIANA
some m: there is a dominated splitting of index p, Tf n x M = En ⊕ Fn ,
n ∈ Z along the orbit of x. Clearly, domination implies that En is the sum
of the Oseledets subspaces of f , at the point f n x, associated to the Lyapunov
exponents λ1 (f, x), . . . , λp (f, x), and Fn is the sum of the spaces associated
to the other exponents. Since this holds whenever λp (f, x) is bigger than
λp+1 (f, x), it proves that the Oseledets splitting is dominated at x.
Theorem 1 is an immediate consequence:
Proof of Theorem 1. The function f → LEj (f ) is semi-continuous for
every j = 1, . . . , d − 1; see Section 2.1.3. Hence, there exists a residual
subset R of Diff 1µ (M ) such that any f ∈ R is a point of continuity for
f → (LE1 (f ), . . . , LEd−1 (f )). By Theorem 2, every point of continuity satisfies the conclusion of Theorem 1.
5. Consequences of nondominance for symplectic maps
Here we prove a symplectic analogue of Proposition 3.1:
Proposition 5.1. Given f ∈ Sympl1ω (M ), ε0 > 0 and 0 < κ < 1, if
m ∈ N is large enough then the following holds:
Let y ∈ M be a nonperiodic point and suppose there exists a nontrivial
splitting Ty M = E ⊕ F into two Lagrangian spaces such that
(5.1)
Dfym |F 1
≥ .
m
m(Dfy |E )
2
Then there exists an (ε0 , κ)-realizable sequence {L0 , . . . , Lm−1 } at y of length m
and there are nonzero vectors v ∈ E, w ∈ Dfym (F ) such that Lm−1 . . . L0 (v) = w.
Remark 5.2. The hypothesis that E and F are Lagrangian subspaces in
Proposition 5.1 is the sole reason why Theorem 4 is weaker than what is stated
in [5].
In subsections 5.1 and 5.2 we prove two results, namely, Lemmas 5.3
and 5.8, that are used in subsection 5.3 to prove Proposition 5.1. In Section 6
we prove Theorems 3 and 4 using Proposition 5.1.
5.1. Symplectic realizable sequences of length 1. First, we recall some
elementary facts: Let (·, ·) denote the usual hermitian inner product in Cq . Up
to identification of Cq with R2q , the standard inner product in R2q is Re(·, ·)
and the standard symplectic form in R2q is Im(·, ·). The unitary group U(q) is
subgroup of GL(q, C) formed by the linear maps that preserve the hermitian
product. When R ∈ U(q) is a map R : R2q → R2q , then R is both symplectic
and orthogonal.
LYAPUNOV EXPONENTS
1469
If R : Tx M → Tx M is an ω-preserving linear map, we shall call R unitary
if it preserves the inner product in Tx M induced from the Euclidean inner
product in R2q by the chart ϕi(x) (recall subsection 2.5).
The next lemma constructs realizable sequences of length 1:
Lemma 5.3. Given f ∈ Sympl1ω (M ), ε0 > 0, κ > 0, there exists ε > 0
with the following properties: Suppose given a nonperiodic point x ∈ M and
a unitary map R : Tx M → Tx M with R − I < ε. Then {Dfx R} is an
(ε0 , κ)-realizable sequence of length 1 at the point x and {R Dff −1 (x) } is an
(ε0 , κ)-realizable sequence of length 1 at the point f −1 (x).
Before starting the proof, let us make some remarks.
Remark 5.4. Let H : R2q → R be a smooth function such that the corresponding Hamiltonian flow ϕt : R2q → R2q is globally defined for every t ∈ R.
= ψ ◦ H. Then the HamilLet ψ : R → R be a smooth function, and define H
t
is globally defined and is given by ϕ̃t (x) = ϕψ (H(x))t (x).
tonian flow (ϕ̃ ) of H
If R ∈ U(q) then all its eigenvalues belong to the unit circle in C. Moreover, there exists an orthonormal basis of Cq formed by eigenvectors of R. If
J ⊂ R is an interval, we define SJ as the set of matrices R ∈ U(q) whose
eigenvalues can be written as eiθ1 , . . . , eiθq , with all θk ∈ J. There is C0 > 0,
depending only on q, such that if ε > 0 and R ∈ S[−ε,ε] then R − I ≤ C0 ε. It
is convenient to consider first the case where the arguments of the eigenvalues
of R all have the same sign and are comparable:
Lemma 5.5. Given ε0 > 0 and 0 < σ < 1, there exists ε > 0 with the
following properties: Given R ∈ S[−3ε,−ε] ∪ S[ε,3ε] , there exists a bounded open
set U ⊂ R2q such that 0 ∈ σU ⊂ U , and there exists a C 1 symplectomorphism
h : R2q → R2q such that
(i) h(z) = z for every z ∈
/ U and h(z) = R(z) for every z ∈ σU ;
(ii) Dhz − I < ε0 for all z ∈ R2q .
Proof.
Let ε0 and σ be given. Let ε > 0 be a small number, to be
specified later. Take R ∈ S[ε,3ε] ; the other possibility is tackled in a similar way.
Let {v1 , . . . , vq } be an orthonormal basis of eigenvectors of R, with associated
eigenvalues eiθ1 , . . . , eiθq , and all ε ≤ θk ≤ 3ε. Up to replacing R with SRS −1 ,
for some S ∈ U(q), we may assume that the basis {v1 , . . . , vq } coincides with
the standard basis of Cq . Therefore R assumes the form
R(z1 , . . . , zq ) = (eiθ1 z1 , . . . , eiθq zq ).
&
Let H : Cq → R be given by H(z) = 12 k θk |zk |2 . Then R is the time 1 map
of the Hamiltonian flow of H. Besides, since max θk ≤ 3 min θk , there is C1
1470
JAIRO BOCHI AND MARCELO VIANA
depending only on q such that
(5.2)
z DHz ≤ C1 H(z)
for all z ∈ Cq .
2
Let τ : R → R be a smooth function such that τ (s) = 1 for
ss ≤ σ , τ (s) = 0
2
for s ≥ 1, and 0 ≤ −τ (s) ≤ 2/(1 − σ ) for all s. Let ψ(s) = 0 τ (u) du and let
= ψ ◦ H. By Remark 5.4, the time 1 map h of the Hamiltonian flow of H
is
H
h(z) = (eiθ1 τ (H(z)) z1 , . . . , eiθk τ (H(z)) zk ).
Then h(z) = R(z) if H(z) ≤ σ 2 and h(z) = z if H(z) ≥ 1. Moreover, a direct
calculation gives
C2 εz DHz Dhz − I ≤
+ 3ε
1 − σ2
where C2 = C2 (q). Due to (5.2), we can take ε = ε(ε0 , σ) such that the righthand side is less than ε0 whenever H(z) < 1. Since H is positive definite, the
set U = {z ∈ Cq ; H(z) < 1} is bounded.
Remark 5.6. We may assume that the set U in Lemma 5.5 has arbitrarily
= aU and h
small diameter. Indeed, if a > 0 then we may replace U with U
−1
with h(z) = ah(a z). Notice that Dhz = Dha−1 z , so that h is a symplectomorphism and satisfies property (ii) of the lemma.
Lemma 5.7. Given ε0 > 0 and 0 < σ < 1, there exists ε > 0 with the
following properties: Given R ∈ S[−ε,ε] , there exists a bounded open set U ⊂
R2q such that 0 ∈ σU ⊂ U , a measurable set K ⊂ U with vol(U K) <
3(1 − σ d ) vol(U ), and a C 1 symplectomorphism h : R2q → R2q such that
(i) h(z) = z for every z ∈
/ U and Dhz = R for every z ∈ K;
(ii) Dhz − I < ε0 for all z ∈ R2q .
Proof. Let ε be as given by Lemma 5.5. Write R ∈ S[−ε,ε] as a product
R = R+ R− , with R+ ∈ S[ε,3ε] and R− = e−2εi I. Applying Lemma 5.5 to R± ,
with ε0 replaced with ε0 /2, we obtain sets U± and symplectomorphisms h± .
Let U = U+ . Consider the family F of all sets of the form aU− + b, with a > 0
and b ∈ R2q , that are contained in U . This is a Vitali covering of U , and so
we may find a finite number of disjoint sets U−i = ai U− + bi ∈ F that cover U
except for a set of volume (1 − σ d ) vol(U ). Using Lemma 5.5 and Remark 5.6,
for each i we find a symplectomorphism hi− such that hi− = id outside U−i and
D(hi− )z = R− for z ∈ Ki = ai σU− + bi , and D(hi− )z is uniformly close to I.
Let K = (σU ) ∩ i K i . Define h = h+ ◦ hi− inside each U−i , and h = h+ outside.
Then K and h have the desired properties.
LYAPUNOV EXPONENTS
1471
Proof of Lemma 5.3. Given ε0 and κ, choose σ close to 1 so that
3(1 − σ d ) < κ. Remark 5.6 also applies to Lemma 5.7: the set U may be
taken with arbitrarily small diameter. Using Lemma 2.13, we conclude that
the sequences {Dfx R} and {R Dff −1 (x) } are (ε0 , κ)-realizable as stated.
5.2. Symplectic nested rotations. In this subsection we prove an analogue
of Lemma 3.3 for symplectic maps:
Lemma 5.8. Given f ∈ Sympl1ω (M ), ε0 > 0, κ > 0, E > 1, and 0 <
γ ≤ π/2, there exists β > 0 with the following properties: Suppose there is a
nonperiodic point x ∈ M , a number n ∈ N, and a two-dimensional symplectic
subspace Y0 ⊂ Tx M such that:
• Df j |Y0 /m(Df j |Y0 ) ≤ E 2 for every j = 1, . . . , n;
• (Xj , Yj ) ≥ γ for each j = 0, . . . , n − 1 where X0 = Y0ω , Xj = Dfxj (X0 ),
and Yj = Dfxj (Y0 ).
Let θ0 , . . . , θn−1 ∈ [−β, β] and let S0 , . . . , Sn−1 : Y0 → Y0 be the rotations of
the plane Y0 by angles θ0 , . . . , θn−1 . Let linear maps
L
L
Ln−1
0
1
Tx M −→
Tf x M −→
. . . −−−→ Tf n (x) M
be defined by Lj (v) = Dff j (x) (v) for v ∈ Xj and Lj (w) = (Dfyj+1 ) · Sj ·
(Dfyj )−1 (w) for w ∈ Yj . Then {L0 , . . . , Ln−1 } is an (ε0 , κ)-realizable sequence
of length n at the point x.
We begin by proving a perturbation lemma that corresponds to Lemma 3.4:
Lemma 5.9. Given ε0 > 0 and 0 < σ < 1, there is ε > 0 with the following
properties: Suppose there exist a splitting R2q = X ⊕ Y with dim Y = 2,
X ω = Y and X ⊥ Y , an ellipsoid A ⊂ X centered at the origin, and a unitary
map R ∈ U(q) with R|X = I and R − I < ε.
Then there exists τ > 1 such that the following holds. Let B be the unit
ball in Y . For a, b > 0 consider the cylinder C = Ca,b = aA ⊕ bB. If a > τ b and
diam C < ε0 then there is a C 1 symplectomorphism h : R2q → R2q satisfying:
(i) h(z) = z for every z ∈
/ C and h(z) = R(z) for every z ∈ σC;
(ii) h(z) − z < ε0 and Dhz − I < ε0 for all z ∈ R2q .
Remark 5.10. If H : R2q → R is a smooth function with bounded DH
and D2 H, then the associated Hamiltonian flow ϕt : R2q → R2q is defined
for every time t ∈ R, and
ϕt (z) − z ≤ |t| sup DH,
for every z ∈ R2q and t ∈ R.
(Dϕt )z − I ≤ exp(|t| sup D2 H) − 1
1472
JAIRO BOCHI AND MARCELO VIANA
Proof of Lemma 5.9. Given √
ε0 and σ, let t̄ > 0 be small, to be specified
later. Let ε > 0 be such that ε < 2 sin t̄. Let X, Y , A, B, and R be as in the
statement. Let A : X → X be a linear map such that A(A) is the unit ball
in X. We define τ = A.
Let H : R2q → R be defined by H(x, y) = H(y) = 12 y2 , where (x, y) are
coordinates with respect to the splitting X ⊕ Y . The Hamiltonian flow of H is
a linear flow (Rt )t , where Rt is√a rotation of angle t in the plane Y , with axis
X. In particular, Rt − I = 2 | sin t| and there exists t0 with |t0 | < t̄ such
that Rt0 = R.
Take numbers a, b > 0 such that a/b > τ and the cylinder C = aA ⊕ bB
has diameter less than ε0 . We are going to construct another Hamiltonian H
which is equal to H inside σC and constant outside C. The symplectomorphism
h will be defined as the time t0 of the Hamiltonian flow associated to H.
For this we need a few auxiliary functions. Let ζ : R → [0, 1] be a smooth
function such that:
• ζ(t) = 1 for t ≤ σ and ζ(t) = 0 for t ≥ 1;
• |ζ (t)| ≤ 10/(1 − σ) and |ζ (t)| ≤ 10/(1 − σ)2 .
Define ψ : X → [0, 1] by ψ(x) = ζ(a−1 Ax). Then
(5.3)
ψ(x) = 1 for x ∈ σaA
and ψ(x) = 0 for x ∈
/ aA.
Let K1 be an upper bound for the norms of the first and second derivatives of
the function x ∈ X → ζ(x). (Notice that K1 depends only on σ.) Then we
have
Dψ ≤ K1 a−1 A and D2 ψ ≤ K1 a−2 A2 .
s
Now define ρ : R → R by ρ(s) = 0 ζ and then let φ : Y → R be given by
φ(y) = 12 b2 ρ(b−1 y)2 . Then
(5.4)
φ(y) = H(y) for y ∈ σbB
and φ(y) = c for y ∈
/ bB,
where 0 < c < 12 b2 is a constant. Besides, we can find K2 > 0, depending only
on σ, such that
Dφ ≤ K2 b and D2 φ ≤ K2 .
: R2q → R by H(x,
y) = c − ψ(x)(c − φ(y)). Then, by (5.3)
Define H
and (5.4),
x ∈ σaA and y ∈ σbB ⇒ H(x,
y) = H(y),
(5.5)
x∈
/ aA or y ∈
/ bB ⇒ H(x,
y) = c.
are (write v = vx + vy ∈ X ⊕ Y and analogously for w)
The derivatives of H
(x,y) (v) = − (c − φ(y))Dψx (vx ) + ψ(x)Dφy (vy ),
DH
(x,y) (v, w) = − (c − φ(y))D2 ψx (vx , wx ) + Dψx (vx )Dφy (wy )
D2 H
+ Dψx (wx )Dφy (vy ) + ψ(x)D2 φy (vy , wy ).
LYAPUNOV EXPONENTS
1473
Using the previous bounds we obtain
≤ 1 K1 b2 a−2 A2 + 2K1 K2 ba−1 A + K2 .
D2 H
2
is bounded by some K that depends
Since a/b > A, we conclude that D2 H
only on σ.
Take h : R2q → R2q to be the time t0 map of the Hamiltonian flow
Property (i) in Lemma 5.9 follows from (5.5). Since diam C
associated to H.
< ε0 , we have h(z) − z < ε0 for all z. By Remark 5.10, Dhz − I ≤
et0 K − 1 ≤ et̄K − 1 < ε0 , if t̄ = t̄(ε0 , σ) is small enough. This proves (ii) and
the lemma.
An ellipse B contained in a 2-dimensional symplectic subspace Y ⊂ R2q
and centered at the origin has eccentricity E if it is the image of the unit ball
under a linear transformation B : Y → Y with B/m(B) = E 2 . If a map
: Y → Y preserves the ellipse B, then B −1 RB
is a rotation of the plane Y of
R
rotates the ellipse B through angle θ.
some angle θ. In this case we say that R
The following statement is a more flexible version of Lemma 5.9. In fact,
it follows from Lemma 5.9 just by a change of the inner product.
Lemma 5.11. Given ε0 > 0, 0 < σ < 1, γ > 0 and E > 1, there is β > 0
with the following properties: Suppose there are given:
• a splitting R2q = X ⊕ Y with dim Y = 2, X ω = Y and (X, Y ) ≥ γ;
• an ellipsoid A ⊂ X centered at the origin;
• an ellipse B ⊂ Y centered at the origin and with eccentricity at most E;
: Y → Y that rotates B through angle θ, with |θ| < β.
• a map R
Then there exists τ > 1 such that the following holds. Let R : R2q → R2q be
the linear map defined by R(v) = v if v ∈ X and R(w) = R(w)
if w ∈ Y . For
a, b > 0 consider the cylinder C = Ca,b = aA ⊕ bB. If a > τ b and diam C < ε0
then there is a C 1 symplectomorphism h : R2q → R2q satisfying:
(i) h(z) = z for every z ∈
/ C and h(z) = R(z) for every z ∈ σC;
(ii) h(z) − z < ε0 and Dhz − I < ε0 for all z ∈ R2q .
Now Lemma 5.8 is proved in the same way as we proved Lemma 3.3, using
Lemmas 5.11 and 3.5. The argument is even a bit simpler since no truncation
(as in Lemma 3.6) is necessary, since we assume that the angles (Xj , Yj ) are
bounded from zero. The details are left to the reader.
5.3. Proof of Proposition 5.1. We use the following lemma, which was also
needed for example 4 in the introduction:
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JAIRO BOCHI AND MARCELO VIANA
Lemma 5.12. Let G ⊂ GL(d, R) be a closed group which acts transitively
in RPd−1 . Then for every ε1 > 0 there exists α > 0 such that if v1 , v2 ∈ RPd−1
satisfy (v1 , v2 ) < α then there exists R ∈ G such that R − I < ε1 and
R(v1 ) = v2 .
Proof. For δ > 0, let Uδ = {R; R ∈ G, R − I < δ}. Given ε > 0,
fix δ > 0 such that if R1 , R2 ∈ Uδ then R2 R1−1 ∈ Uε1 . The hypothesis on
the group implies that for any w ∈ RPd−1 , the map G → RPd−1 given by
A → A(w) is open (this follows from [15, Th. II.3.2]). Therefore, for any
δ > 0, the set Uδ (w) = {Rw; R ∈ Uδ } is an open neighborhood of w. Cover
RPd−1 by some finite union Uδ (w1 ) ∪ · · · ∪ Uδ (wk ). Now take two directions v1 ,
v2 ∈ RPd−1 sufficiently close. Then both belong to some Uδ (wi ), and so there
are R1 , R2 ∈ Uδ such that v1 = R1 wi and v2 = R2 wi . Therefore R = R2 R1−1
belongs to Uε1 and Rv1 = v2 .
Proof of Proposition 5.1. Let f , ε0 , κ be given. Fix 0 < κ < 12 κ.
Let ε > 0, depending on f , ε0 , κ , be given by Lemma 5.3. Let α > 0,
depending on ε1 = ε and G = U(q), be given by Lemma 5.12. Take K
satisfying K ≥ (sin α)−2 and K ≥ maxx Dfx /m(Dfx ). Let E > 1 and γ > 0
be given by
E 2 = 8Cω4 K(sin α)−4
and
sin γ = 12 Cω−14 K −2 sin9 α,
where Cω is as in (2.4). Let β > 0 be given by Lemma 5.8. Finally, let
m ≥ 2π/β. The proof is divided into three cases.
First case. Suppose that there exists ∈ {0, 1, . . . , m} such that
(5.6)
(E , F ) < α.
Fix as above and take unit vectors ξ ∈ E , η ∈ F such that (ξ, η) < α.
By Lemma 5.12, there exists a unitary transformation R : Tf (y) M → Tf (y) M
such that R−I < ε and R(ξ) = η. By Lemma 5.3, the sequences {Dff (x) R}
and {R Dff −1 (x) } are (κ , ε0 )-realizable. Define {L0 , . . . Lm−1 } as
{Dfy , . . . , Dff −1 (y) , Dff (y) R, Dff +1 (y) , . . . , Dff m−1 (y) }
if < m and as {Dfy , . . . , Dff m−2 (y) , R Dff m−1 (y) } if = m. In either case,
this is a (κ, ε0 )-realizable sequence of length m at y, whose product Lm−1 . . . L0
sends the direction RDf − (ξ) ⊂ E0 to the direction RDf m− (η) ⊂ Fm .
Second case. Assume that there exist k, ∈ {0, . . . , m} with k < and
(5.7)
Dff−k
k (y) |Fk m(Dff−k
k (y) |Ek )
> K.
The proof of this case is easily adapted from the second case in the proof of
Proposition 3.1. We leave the details to the reader.
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LYAPUNOV EXPONENTS
Third case. We suppose that we are not in the previous cases, that is,
for every j ∈ {0, 1, . . . , m}, (Ej , Fj ) ≥ α,
(5.8)
and
for every i, j ∈ {0, . . . , m} with i < j,
(5.9)
Dffj−i
i (y) |Fi m(Dffj−i
i (y) |Ei )
≤ K.
By (5.8) and Lemma 2.3, we have, for all i, j ∈ {0, . . . , m} with i < j,
Cω−2 sin α ≤ m(Df j−i |Ei ) Df j−i |Fi ≤ Cω2 (sin α)−1 .
(5.10)
This, together with (5.9), gives
(5.11)
m(Df j−i |Ei ) ≥ Cω−1 K −1/2 (sin α)1/2 ,
(5.12)
Df j−i |Fi ≤ Cω K 1/2 (sin α)−1/2 .
Also, by (5.10) and the main assumption (5.1),
m(Df m |E0 ) ≤ 21/2 Cω (sin α)−1/2 ,
(5.13)
Df m |F0 ≥ 2−1/2 Cω−1 (sin α)1/2 .
(5.14)
Let v0 ∈ E0 be such that v0 = 1 and Dfym (v0 ) = m(Dfym |E0 ). Using
Lemma 2.3.1, take w0 ∈ F0 with w0 = 1 such that |ω(v0 , w0 )| ≥ Cω−1 sin α.
Let G0 = E0 ∩ w0ω and H0 = F0 ∩ v0ω . (By v ω we mean (Rv)ω .) Notice that
E0 = Rv0 ⊕ G0 and F0 = Rw0 ⊕ H0 . Let X0 = G0 ⊕ H0 and Y0 = Rv0 ⊕ Rw0 .
Then X0 = Y0ω . Let, for j = 1, . . . , m,
vj = Df j (v0 )/Df j (v0 ),
Gj = Df j (G0 ),
Xj = Df j (X0 ),
wj = Df j (w0 )/Df j (w0 ),
Hj = Df j (H0 ),
Yj = Df j (Y0 )
(all the derivatives are at y). By (2.5),
Cω−1 sin α ≤ |ω(v0 , w0 )| = |ω(Df m v0 , Df m w0 )| ≤ Cω Df m v0 Df m w0 .
Thus Df m w0 ≥ Cω−2 sin α · m(Df m |E0 )−1 and, by (5.10),
Df m w0 ≥ Cω−4 sin2 α · Df m |F0 ;
(5.15)
that is, w0 is “almost” the most expanded vector by Df m in F0 . Hence,
by (5.1),
Df m w0 Df m |F0 ≥ Cω−4 sin2 α
≥ 12 Cω−4 sin2 α.
m
Df v0 m(Df m |E0 )
This and (5.9) imply that, for each j = 1, . . . , m,
K≥
Df m w0 /Df m−j |Fj Df j w0 ≥
≥ 12 Cω−4 K −1 sin2 α.
Df j v0 Df m v0 /m(Df m−j |Ej )
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JAIRO BOCHI AND MARCELO VIANA
Therefore, by (5.8) and Lemma 2.8,
Df j |Y0 ≤ 8Cω4 K(sin α)−4 = E 2 .
m(Df j |Y0 )
(5.16)
We now deduce some angle estimates. First, we claim that
(5.17)
sin (v0 , G0 ) ≥ Cω−2 sin α
and
sin (w0 , H0 ) ≥ Cω−2 sin α.
Indeed, write v0 = u+u with u ∈ G0 and u ⊥ G0 . Since G0 is skew-orthogonal
to w0 ,
Cω−1 sin α ≤ |ω(v0 , w0 )| = |ω(u, w0 )| ≤ Cω u.
That is, sin (v0 , G0 ) = u ≥ Cω−2 sin α. Analogously we prove the other
inequality in (5.17). Next, we estimate sin (vj , Gj ) and sin (wj , Hj ) for
j = 1, . . . , m. For this we use relation (2.6) from subsection 2.4, which gives:
m(Df j |E0 )
sin (v0 , G0 ),
Df j v0 Df j w0 sin (w0 , H0 ).
sin (wj , Hj ) ≥
Df j |F0 sin (vj , Gj ) ≥
(5.18)
(5.19)
By (5.11) and (5.13),
Df j v0 =
Df m v0 m(Df m |E0 )
≤
≤ 21/2 Cω2 K 1/2 (sin α)−1
Df m−j vj m(Df m−j |Ej )
for each j = 1, . . . , m. So, by (5.11) again,
Df j v0 ≤ 21/2 Cω3 K(sin α)−3/2 .
m(Df j |E0 )
This, together with (5.17) and (5.18), gives
(5.20)
sin (vj , Gj ) ≥ 2−1/2 Cω−5 K −1 (sin α)5/2 .
Similarly, by (5.15), (5.12), and (5.14),
Df j w0 =
Df m w0 Df m |F0 −4
2
sin
α
≥
C
≥ 2−1/2 Cω−6 K −1/2 sin3 α.
ω
Df m−j wj Df m−j |F0 Then, using (5.12) again, we get
Df j w0 ≥ 2−1/2 Cω−7 K −1 (sin α)7/2 .
Df j |F0 By (5.17) and (5.19),
(5.21)
sin (wj , Hj ) ≥ 2−1/2 Cω−9 K −1 (sin α)9/2 .
LYAPUNOV EXPONENTS
1477
Now we use Lemma 2.6 three times:
sin (Yj , Xj ) ≥ sin (vj , Xj ) sin (wj , Rvj ⊕ Xj )
≥ sin (vj , Gj ) sin (Ej , Hj ) sin (wj , Rvj ⊕ Xj )
≥ sin (vj , Gj ) sin (wj , Hj ) sin2 (Ej , Hj )
So, using (5.20), (5.21), and (Ej , Hj ) ≥ α, we obtain
(5.22)
sin (Xj , Yj ) ≥ 12 Cω−14 K −2 sin9 α = sin γ.
Relations (5.16) and (5.22) permit us to apply Lemma 5.8. Since mβ ≥ 2π,
&
θj =
it is possible to choose numbers θ0 , . . . , θm−1 such that 0 ≤ θj ≤ β and
(v0 , w0 ). Let Sj and Lj be as in Lemma 5.8. We have Lm−1 . . . L0 |Y0 =
(Df m |Y0 )Sm−1 . . . S0 , so that Lm−1 . . . L0 (Rv0 ) = Rwm . This completes the
proof of Proposition 5.1.
6. Proofs of Theorems 3 and 4
Given f ∈ Diff 1µ (M ) and m ∈ N, let D(f, m) be the (closed) set of points
x such that there is an m-dominated splitting of index q = d/2 along the
orbit of x. Let Γ(f, m) = M D(f, m) and let Γ∗ (f, m) be the set of points
x ∈ Γ(f, m) which are regular, not periodic, and satisfy λq (f, x) > 0. Let also
Γ(f, ∞) = m∈N Γ(f, m).
We have the following symplectic analogues of Propositions 4.2, 4.8 and
4.17:
Proposition 6.1. Let f ∈ Sympl1ω (M ), ε0 > 0, δ > 0, and 0 < κ < 1.
If m ∈ N is sufficiently large, then there exists a measurable function N :
Γ∗ (f, m) → N such that for a.e. x ∈ Γ∗ (f, m) and every n ≥ N (x) there exists
0 , . . . , L
n−1 } at x of length n such that
an (ε0 , κ)-realizable sequence {L
1
n−1 . . . L
0 ) ≤ Λq−1 (f, x) + δ.
log ∧q (L
n
Proposition 6.2. Let f ∈ Sympl1ω (M ), ε0 > 0 and δ > 0. Then there
exist m ∈ N and a diffeomorphism g ∈ U(f, ε0 ) that equals f outside the open
set Γ(f, m) and such that
Λq (g, x) dµ(x) < δ +
Λq−1 (f, x) dµ(x).
Γ(f,m)
Γ(f,m)
Proposition 6.3. Given f ∈ Sympl1ω (M ), let
λq (f, x) dµ(x).
J(f ) =
Γ(f,∞)
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JAIRO BOCHI AND MARCELO VIANA
Then for every ε0 > 0 and δ > 0, there exists a diffeomorphism g ∈ U(f, ε0 )
such that
Λq (g, x) dµ(x) <
Λq (f, x) dµ(x) − J(f ) + δ.
M
M
The proofs of these propositions are exactly the same as those of the
corresponding results in Section 4, in the following logical order:
Proposition 5.1 ⇒ Proposition 6.1 ⇒ Proposition 6.2 ⇒ Proposition 6.3.
Concerning the first implication, notice that if x ∈ Γ∗ (f, m) then, by Lemma 2.4,
the spaces Ex+ and Ex− (that correspond to positive and negative Lyapunov exponents) are Lagrangian, so that Proposition 5.1 applies.
6.1. Conclusion of the proofs of Theorems 3 and 4.
Proof of Theorem 3. Let f ∈ Sympl1ω (M ) be a point of continuity of
the map LEq (·). By Proposition 6.3, J(f ) = 0, that is, λq (f, x) = 0 for
a.e. x ∈ Γ(f, ∞). Let x ∈ M be a regular point. If λq (f, x) > 0, we have
(if we exclude a zero measure set of x) x ∈
/ Γ(f, ∞). This means that there
is a dominated splitting, Tf n (x) M = En ⊕ Fn , n ∈ Z of index q, along the
orbit of x. Then En is the sum of the Oseledets spaces of f , at the point f n x,
associated to the Lyapunov exponents λ1 (f, x), . . . , λq (f, x), and Fn is the
sum of the spaces associated to the other exponents. By part 2 of Lemma 2.4,
the splitting Tf n (x) M = En ⊕ Fn , n ∈ Z, is hyperbolic.
The next proposition is used to deduce Theorem 4 from Theorem 3.
Proposition 6.4. There is a residual subset R2 ⊂ Sympl1ω (M ) such that
if f ∈ R2 then either f is Anosov or every hyperbolic set of f has measure 0.
Proof. This is a modification of an argument from [20]. We use the fact,
proved in [30], that C 2 diffeomorphisms are dense in the space Sympl1ω (M ).
Another key ingredient is that the hyperbolic sets of any C 2 non-Anosov diffeomorphism have zero measure. We comment on the latter near the end.
For each open set U ⊂ M with U = M and each f ∈ Sympl1ω (M ), consider
the maximal f -invariant set inside U ,
f n (U ).
Λf (U ) =
n∈Z
For ε > 0, let D(ε, U ) be the set of diffeomorphisms f ∈ Sympl1ω (M ) such that
at least one of the following properties is satisfied:
(i) There is a neighborhood U of f such that Λg (U ) is not hyperbolic for all
g ∈ U;
(ii) µ(Λf (U )) < ε.
LYAPUNOV EXPONENTS
1479
Clearly, the set D(ε, U ) is open. Moreover, it is dense. Indeed, if f does not
satisfy (i) then there is g close to f such that Λg (U ) is hyperbolic. Take f1 ∈ C 2
close to g in Sympl1ω (M ). Then Λf1 (U ) is hyperbolic with measure zero, and
so f1 ∈ D(ε, U ). This proves denseness. Hence the set
D(U ) = ∩ε>0 D(ε, U )
⊃ {f ∈ Sympl1ω (M ); Λf (U ) is hyperbolic ⇒ µ(Λf (U )) = 0}
is residual. Now take B a countable basis of open sets of M and let B be the
set of all finite unions of sets in B. The set
R2 =
D(U )
U =M
U ∈B,
is residual in Sympl1ω (M ) and the hyperbolic sets for every non-Anosov f ∈ R
have zero measure.
Finally, we explain why all hyperbolic sets of a C 2 non-Anosov diffeomorphism have zero measure. This is well-known for hyperbolic basic sets; see [11].
We just outline the arguments in the general case. Suppose f has a hyperbolic
set Λ with µ(Λ) > 0. Using absolute continuity of the unstable lamination,
we get that µu (Wεu (x) ∩ Λ) > 0 for some x ∈ Λ, where µu denotes Lebesgue
measure along unstable manifolds. By bounded distortion and a density point
argument, we find points xk ∈ Λ such that µu (Wεu (xk ) Λ) converges to zero.
Taking an accumulation point x0 we get that Wεu (x0 ) ⊂ Λ. We may suppose
that every point of Λ is in the support of µ|Λ. In particular, there are recurrent
points of Λ close to x0 . Applying the shadowing lemma, we find a hyperbolic
periodic point p0 close to x0 . In particular, Wεs (p0 ) intersects Wεu (x0 ) transversely. Using the λ-lemma we conclude that the whole W u (p0 ) is contained
in Λ. Define Λ0 as the closure of the unstable manifold of the orbit of p0 . This
is a hyperbolic set contained in Λ, and it consists of entire unstable manifolds.
Hence, W s (Λ0 ) is an open neighborhood of Λ0 . Using the fact that f preserves
volume, we check that f (Wεs (Λ0 )) = Wεs (Λ0 ). This implies that W s (Λ0 ) = Λ0
and so, by connectedness, Λ0 must be the whole M . Consequently, f is Anosov.
Proof of Theorem 4. It suffices to take R = R1 ∩ R2 with R1 a residual
set of continuity points of f → LEq (f ), and R2 as in Proposition 6.4.
7. Proof of Theorem 5
Let M be a compact Hausdorff space, µ a Borel regular measure and f :
M → M a homeomorphism preserving the measure µ. Let S be an accessible
subset of GL(d, R), according to Definition 1.2.
The following result provides an analogue of Proposition 3.1:
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JAIRO BOCHI AND MARCELO VIANA
Proposition 7.1. Given A ∈ C(M, S) and ε > 0, if m
∈ N is large
enough then the following holds:
Let y ∈ M be a nonperiodic point and suppose it is given a nontrivial
splitting Rd = E ⊕ F such that
Am
(y)|F 1
≥ .
m
2
m(A (y)|E )
(7.1)
∈ S, with Lj − A(f j y) < ε, such
Then there exist matrices L0 , . . . , Lm−1
that Lm−1
. . . L0 (v) = w for some nonzero vectors v ∈ E and w ∈ Am
(y)(F ).
Proof. Let C0 be such that A(x)±1 ≤ C0 for all x ∈ M . Let ν ∈ N and
α > 0, depending on C0 and ε, be as in Definition 1.2. Let
K = max{(sin α)−1 , C02 },
and C = 8C02 K(sin α)−2 .
Given m
large, let m ∈ N be such that 1 ≤ m
− νm ≤ ν. We assume m
is
large enough so that m > 2C/α.
Now take y, E and F as in the statement. For j = 0, 1, . . . , m − 1, let
Aj = A(f νj y), Ej = Aνj (x)(E), Fj = Aνj (x)(F ). (We disregard times which
are not multiples of ν.) As before, we divide the rest of the proof into three
cases:
First case. There exists ∈ {0, . . . , m} such that
(E , F ) < α.
(7.2)
With as above, we take ξ ∈ E , η ∈ F such that (ξ, η) < α. By definiν−1 such that A
i −A(f ν+i x) < ε and
0 , . . . , A
tion of accessibility, there are A
0 (Rξ) = Aν (f ν y)(Rη). We define Lj = A
j−ν for ν
≤ j < ν(
+ 1)
ν−1 . . . A
A
j
and Lj = A(f y) for the remaining j’s. Then {L0 , . . . , Lm−1
} has the required
properties.
Second case.
Assume that there exist k, ∈ {0, . . . , m} such that k < and
A−1 . . . Ak |Fk > K.
m(A−1 . . . Ak |Ek )
Once more, this is similar to the second case in Propositions 3.1 and 5.1. We
leave it to the reader to spell-out the details.
(7.3)
Third case.
we assume
(7.4)
We suppose that we are not in the previous cases, that is,
for every j ∈ {0, 1, . . . , m}, (Ej , Fj ) ≥ α,
and
(7.5)
for every i, j ∈ {0, . . . , m} with i < j,
Aj−1 . . . Ai |Fi ≤ K.
m(Aj−1 . . . Ai |Ei )
LYAPUNOV EXPONENTS
1481
Take unit vectors ξ ∈ E0 and η ∈ F0 such that
Am−1 . . . A0 (ξ) = m(Am−1 . . . A0 |E0 )
and
Am−1 . . . A0 (η) = Am−1 . . . A0 |F0 .
Let ξj = Aj−1 . . . A0 (ξ), ηj = Aj−1 . . . A0 (η) and Yj = Rξj ⊕ Rηj .
The assumption (7.1) gives
(y)|F / Am−νm
(y)|F Am−1 . . . A0 (η)
Am
1
.
≥
≥
m
m−νm
Am−1 . . . A0 (ξ)
m(A (y)|E ) / m(A
(y)|E )
2C02ν
By (7.5), for any j = 1, . . . , m,
K≥
Aj−1 . . . A0 (η)
Am−1 . . . A0 (η) / Am−1 . . . Aj 1
.
≥
≥
Aj−1 . . . A0 (ξ)
Am−1 . . . A0 (ξ) / m(Am−1 . . . Aj )
2C02ν K
This, together with (7.4) and Lemma 2.8 implies that, for all j = 1, . . . , m,
Aj−1 . . . A0 |Y0 ≤ C.
m(Aj−1 . . . A0 |Y0 )
(7.6)
Assign orientations to the planes Yj such that each Aj |Yj : Yj → Yj+1
preserves orientation. Let Pj be the projective space of Yj , with the induced
orientation. Let vj = Rξj and wj = Rηj ∈ Pj . For each z ∈ Pj , let [z] ∈ [0, π)
be the oriented angle between z and vj . Now, z → [z] is a bijection and
[z] → [Aj z] is monotonic. If L : Y0 → Yj is any linear map then, by Lemma 2.7,
π
[Lz2 ] − [Lz1 ]
2 L
=⇒
≤ ·
.
2
[z2 ] − [z1 ]
π m(L)
We define directions u0 ∈ P0 , . . . , um ∈ Pm by recurrence as follows: Let
[u0 ] = 0 and
(7.7)
0 < [z2 ] − [z1 ] ≤
(7.8)
[uj+1 ] = min{[wj+1 ], [Aj (uj + α)]}.
We claim that [um ] = [wm ]. Indeed, [Aj uj ] ≤ [uj+1 ] ≤ [wj+1 ] for each j < m.
Therefore, defining [zj ] = [(Aj−1 . . . A0 )−1 uj ], we have
0 = [z0 ] ≤ [z1 ] ≤ · · · ≤ [zm ] ≤ [w0 ] < π.
In particular, [zi+1 ] − [zi ] < π/m for some i = 0, . . . , m − 1. Hence, by (7.6)
and (7.7),
[A−1
i ui+1 ] − [ui ] = [Ai−1 . . . A0 zi+1 ] − [Ai−1 . . . A0 zi ] < 2C/m ≤ α.
Due to (7.8), this is only possible if [ui+1 ] = [wi+1 ]. This implies [um ] = [wm ].
j,0 , . . . , A
j,ν−1 ∈ S
Now, (A−1 uj+1 , uj ) < α for each j, so that we can find A
j
j,k − A(f νj+k x) < ε and A
j,ν−1 . . . A
j,0 (Ruj ) = Ruj+1 . We define
such that A
the sequence {L0 , . . . , Lm−1
} as
0,0 , . . . , A
j,ν−1 , . . . , A
m−1,0 , . . . , A
m−1,ν−1 , A(f νm x), . . . , A(f m−1
{A
x)}.
. . . L0 (v0 ) ∈ Am
(y)(F ).
Then Lm−1
1482
JAIRO BOCHI AND MARCELO VIANA
Next we define sets Γp (A, m), Γ∗p (A, m), Γp (A, m) for p ∈ {1, . . . , d − 1}
and m ∈ N, in the same way as in Section 4, with the obvious adaptations.
Lemma 4.1 also applies in the present context.
Proposition 7.2. Given A ∈ C(M, S), ε > 0, δ > 0, and p ∈ {1, . . . ,
d − 1}, if m ∈ N is sufficiently large then there exists a measurable function
N : Γ∗p (A, m) → N such that for a.e. x ∈ Γ∗p (A, m) and every n ≥ N (x) there
0 , . . . , L
n−1 ∈ S such that L
j − A(f j x) < ε and
exist matrices L
1
n−1 . . . L
0 ) ≤ Λp−1 (A, x) + Λp+1 (A, x) + δ.
log ∧p (L
n
2
The proof is the same as that for Proposition 4.2.
Proposition 7.3. Let A ∈ C(M, S), ε0 > 0, p ∈ {1, . . . , d−1} and δ > 0.
Then there exist m ∈ N and a cocycle B ∈ C(M, S), with B − A∞ < ε0 , that
equals A outside the open set Γp (A, m) and is such that
Λp (B, x) dµ(x) < δ +
Γp (A,m)
Γp (A,m)
Λp−1 (A, x) + Λp+1 (A, x)
dµ(x).
2
The proof of Proposition 7.3 is not just an adaptation of that of Proposition 4.8, because Vitali’s lemma may not apply to M . We begin by proving
a weaker statement, in Lemma 7.4. Let L∞ (M, S) denote the set of bounded
measurable functions from M to S. Oseledets’ theorem also applies for cocycles
in L∞ (M, S).
Lemma 7.4. Let A ∈ C(M, S), with ε0 > 0, p ∈ {1, . . . , d − 1} and δ > 0.
∈ L∞ (M, S), with B
− A∞ < ε0 /2,
Then there exist m ∈ N and a cocycle B
that equals A outside the open set Γp (A, m) and such that
Γp (A,m)
x) dµ(x) < δ +
Λp (B,
Γp (A,m)
Λp−1 (A, x) + Λp+1 (A, x)
dµ(x).
2
Sketch of proof.
We shall explain the necessary modifications of the
i and Qi are defined as before. In
proof of Proposition 4.8. The sets Z i , Q
Lemma 4.14, the castles U i and K i become equal to Qi (as κ and γ were 0).
We decompose each base Qib into finitely many disjoint measurable sets Uki with
small diameter. In each tower with base Uki we construct the perturbation B
constant in each floor. The definitions of N
using Proposition 7.2, taking B
i
and G are the same. In Lemma 4.16 several bounds (those involving κ or γ)
become trivial. Then one concludes the proof in the same way as before.
LYAPUNOV EXPONENTS
1483
Proof of Proposition 7.3. Let A, ε0 , p and δ be as in the statement. Let
be given by Lemma 7.4. Let N ∈ N be such that
m and B
Λp−1 (A, x) + Λp+1 (A, x)
1
N (x)) dµ < 2δ +
log ∧p (B
dµ.
2
Γp (A,m) N
Γp (A,m)
Let γ = N −1 δ. Using Lusin’s theorem (see [28]) and the fact that S is a
= A outmanifold, one finds a continuous B : M → S such that B = B
∞ < ε0 /2, and the set E =
side the open set Γp (A, m), the norm B − B
{x ∈ M ; B(x) = B(x)} has measure µ(E) < γ. Let
G=
N
−1
f −j Γp (A, m) E ⊂ Γp (A, m).
j=0
Then µ Γp (A, m) G ≤ N µ(E) < δ. Now, letting C be an upper bound for
we have
log ∧p (B(x)),
1
Λp (B, x) dµ ≤
log ∧p (B N (x)) dµ
Γp (A,m)
Γp (A,m) N
Λp−1 (A, x) + Λp+1 (A, x)
dµ.
< Cδ + 2δ +
2
Γp (A,m)
Up to replacement of δ with δ/(C + 2), this completes the proof.
Using Proposition 7.3, one concludes the proof of Theorem 5 exactly as in
subsection 4.3. The fact that either vanishing of the exponents or dominance of
the splitting is also a sufficient condition for continuity is an easy consequence of
semi-continuity of Lyapunov exponents and robustness of dominated splittings
under small perturbations of the cocycle.
IMPA, Rio de Janeiro, Brazil
Current address: Instituto de Matemática, UFRGS, Porto Alegre, Brazil
E-mail address: [email protected]
IMPA, Rio de Janeiro, Brazil
E-mail address: [email protected]
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(Received September 11, 2002)